{"id":88292,"date":"2021-06-24T16:51:00","date_gmt":"2021-06-24T11:21:00","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=88292"},"modified":"2021-06-30T16:33:41","modified_gmt":"2021-06-30T11:03:41","slug":"1st-puc-electronics-question-bank-chapter-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-electronics-question-bank-chapter-2\/","title":{"rendered":"1st PUC Electronics Question Bank Chapter 2 Principles of Electricity, Network Theorems and AC Principles"},"content":{"rendered":"

Karnataka 1st PUC Electronics Question Bank Chapter 2 Principles of Electricity, Network Theorems and AC Principles<\/h2>\n

One Mark Questions and Answers<\/span><\/p>\n

Question 1.
\nWhat is the unit of electric charge?
\nAnswer:
\ncoulomb (C).<\/p>\n

Question 2.
\nWhat is the magnitude of the charge of an electron?
\nAnswer:
\n1.6 \u00d7 10-19<\/sup> C<\/p>\n

\"1st<\/p>\n

Question 3.
\nDefine potential differences.
\nThe potential difference between any two points in a circuit is the amount of work done in moving unit positive charge from one point to another.<\/p>\n

Question 4.
\nDefine electric current.
\nElectric current is defined as the rate of flow of charge through a cross-section of a conductor.<\/p>\n

Question 5.
\nWhat is the unit of electric current?
\nampere (A).<\/p>\n

Question 6.
\nWhat is the direction of conventional current?
\nThe direction of conventional current is opposite to the direction of the flow of electrons.<\/p>\n

Question 7.
\nDefine DC current.
\nDC current is the current that flows only in one direction and its magnitude remains constant with time.<\/p>\n

Question 8.
\nDefine ampere.
\nThe current flowing in a conductor is said to be one ampere if one coulomb of charge flows through the conductor in one second.<\/p>\n

Question 9.
\nGive an example for the DC source.
\nA battery.<\/p>\n

Question 10.
\nState ohms law.
\nOhms law states that current flowing through a conductor is directly proportional to the potential difference across the conductor when the temperature and other physical conditions of the conductor are maintained constant.<\/p>\n

Question 11.
\nIs ohms law applicable when the temperature of a conductor continuously changes?
\nNo.<\/p>\n

\"1st<\/p>\n

Question 12.
\nIs ohm\u2019s law applicable to semiconductors and insulators?
\nNo.<\/p>\n

Question 13.
\nWhat is the resistance of a conductor?
\nAns. Resistance of a conductor is the opposition offered by it to the flow of current.<\/p>\n

Question 14.
\nMention the unit of resistance.
\nOhm (Q).<\/p>\n

Question 15.
\nWhat is a node?
\nAns.
\nNode is a point in a circuit where more than two resistors meet.<\/p>\n

Question 16.
\nWhat is a loop in an electrical circuit?
\nAnswer:
\nA loop is any closed path in an electrical circuit<\/p>\n

Question 17.
\nWhen will a load receive maximum power from a source?
\nAnswer:
\nLoad receives maximum power from a source when the load resistance is equal to source resistance.<\/p>\n

Question 18.
\nHow is Thevenin’s resistance to a network determined?
\nAnswer:
\nThe venins resistance is the resistance between the open terminals with all sources replaced by their internal resistances.<\/p>\n

Question 19.
\nWrite an expression for the instantaneous value of AC Voltage.
\nAnswer:
\nInstantaneous voltage V = Vm<\/sub> sin \u03c9t.<\/p>\n

\"1st<\/p>\n

Question 20.
\nHow much is the AC voltage applied to all household electrical devices?
\nAnswer:
\n230V.<\/p>\n

Two Mark Questions and Answers<\/span><\/p>\n

Question 1.
\nMention any two properties of charges.
\nAnswer:<\/p>\n

    \n
  1. I like charges repel, unlike charges attract.<\/li>\n
  2. Charges always reside on the outer surface of a conductor only.<\/li>\n<\/ol>\n

    Question 2.
    \nWhat is the difference between conventional current and electron current?
    \nAnswer:
    \nConventional current has a direction opposite to that of the flow of electrons. Electron current is in the direction of the flow of electrons.<\/p>\n

    Question 3.
    \nMention limitations of Ohms law.
    \nAnswer:<\/p>\n

      \n
    1. Ohms law is not valid when the physical conditions of the conductor change,<\/li>\n
    2. Ohms law is not valid at very low and very high temperatures.<\/li>\n<\/ol>\n

      Question 4.
      \nWhat are primary sources? Give one example.
      \nAnswer:
      \nPrimary sources are those which can not be recharged once it is completely discharged, eg., alkaline batteries.<\/p>\n

      Question 5.
      \nWhat are secondary DC sources? Give an example.
      \nAnswer:
      \nSecondary sources are those which can be recharged, eg lithium-ion batteries used in mobile phones.<\/p>\n

      \"1st<\/p>\n

      Question 6.
      \nState Kirchhoffs laws.
      \nAnswer:<\/p>\n

        \n
      • Kirchhoff\u2019s current law: The sum of currents towards a node is equal to the sum of the currents away from the node.<\/li>\n
      • Kirchhoff\u2019s voltage law: The algebraic sums of EMFs in any closed loop of a network are equal to the algebraic sum of IR drops in that loop.<\/li>\n<\/ul>\n

        Question 7.
        \nWhat is (i) series combination (ii) parallel combination of resistors?
        \nAnswer:<\/p>\n

          \n
        1. Series combination is connecting the end of the resistor to the end so that the same current flows; through all of them.<\/li>\n
        2. A parallel combination is connecting the resistors between the same two common points.<\/li>\n<\/ol>\n

          Question 8.
          \nDefine peak value and RMS value of ac.
          \nAnswer:
          \nThe peak or maximum value of ac is the maximum voltage or current attained by AC in the half-cycle.
          \nRMS value of ac is the equivalent steady DC value that gives the same heating effect at the same rate in a given resistor.<\/p>\n

          \"1st<\/p>\n

          Question 9.
          \nDefine (i) average value (ii) time period of ac.
          \nAnswer:<\/p>\n

            \n
          1. Average the mean value of ac is the average of instantaneous voltages of ac taken over the one-half cycle.<\/li>\n
          2. The time period of ac is the time taken to complete one AC cycle.<\/li>\n<\/ol>\n

            Question 10.
            \nWhat is the charge on the body to which 1030 electrons are added?
            \nAnswer:
            \nCharge, Q = -ne = -1030<\/sup> \u00d7 1.6 \u00d7 10-19<\/sup> = -1.6 \u00d7 1011<\/sup> C.<\/p>\n

            Question 11.
            \nHow many electrons are to be removed from a body to charge it to 16nC?
            \nAnswer:
            \nn = \\(\\frac{\\mathrm{Q}}{\\mathrm{e}}=\\frac{16 \\times 10^{-9}}{1.6 \\times 10^{-19}}\\) = 10 \u00d7 1010<\/sup> electrons<\/p>\n

            Question 12.
            \nWhat are the resistances offered by (i) short circuit (ii) Open circuit?
            \nAnswer:<\/p>\n

              \n
            1. Zero<\/li>\n
            2. Infinity.<\/li>\n<\/ol>\n

              \"1st<\/p>\n

              Question 13.
              \nDraw any two non-sinusoidal AC waveforms.
              \nAnswer:
              \n\"1st\"1st<\/p>\n

              Three Mark Questions and Answers<\/span><\/p>\n

              Question 1.
              \nWrite any three limitations of Ohms law.
              \nAnswer:<\/p>\n

                \n
              1. Ohm\u2019s law is not valid at very low and very high temperatures.<\/li>\n
              2. When the physical conditions of a conductor change, Ohms law is not valid.<\/li>\n
              3. Ohm\u2019s law is not applicable to semiconductors, discharge tubes, and electrolytes.<\/li>\n<\/ol>\n

                Question 2.
                \nState Kirchhoffs laws.
                \nAnswer:<\/p>\n

                  \n
                1. The sum of the currents flowing towards a node in a circuit is equal to the sum of the currents away from the node.<\/li>\n
                2. The algebraic sum of EMFs in any closed loop of a network is equal to the algebraic sum of IR drops in that loop.<\/li>\n<\/ol>\n

                  Question 3.
                  \nWhat is a voltage source? The row is converted into a current source?
                  \nAnswer:
                  \nThe voltage source is that which supplies electrical energy to the circuit in the form of voltage.<\/p>\n

                  RMS value of ac is the equivalent steady DC value which gives the same heating effect at the same rate in a given resistor.
                  \n\"1st<\/p>\n

                  Question 4.
                  \nWhat is a current source? How is it converted into a voltage source?
                  \nAnswer:
                  \nThe current source is that which supplies electrical energy to the circuit in the form of current.
                  \n\"1st<\/p>\n

                  Question 5.
                  \nWhat are linear and nonlinear circuits? Mention one example for each?
                  \nAnswer:<\/p>\n

                    \n
                  • Linear Network: In a linear circuit, current and voltage are proportionately varying, an eg. circuit having resistor.<\/li>\n
                  • Non-linear Network: In this circuit, current and voltage are not proportionately varying with each other, eg. circuit having a diode or a transistor.<\/li>\n<\/ul>\n

                    Question 6.
                    \nMention unit of frequency. Find frequency of an AC whose time period is 10 ms.
                    \nAnswer:
                    \nFrequency of ac is the number of AC cycles completed in one second.
                    \nf = \\(\\frac{1}{T}=\\frac{1}{10 \\times 10^{-3}}=\\frac{1000}{10}\\) = 100Hz<\/p>\n

                    Question 7.
                    \nWhat are ideal voltage and current sources? Draw the I-V characteristics of a practical voltage source.
                    \nAnswer:
                    \nThe ideal voltage source is the one that supplies a constant voltage to the load irrespective of changes in the load resistance. It has zero resistance. The ideal current source is that which supplies a constant current to the load irrespective of changes in the load. It has infinite resistance.
                    \n\"1st
                    \nV-I characteristics of Practical voltage Source.<\/p>\n

                    \"1st<\/p>\n

                    Question 8.
                    \nFor the AC waveform given below, find (i) Vrms (ii) Vp (iii) Vavg (IV) frequency.
                    \nAnswer:
                    \n\"1st<\/p>\n

                      \n
                    1. Vp = 2v<\/li>\n
                    2. Vrms = \\(\\frac{\\mathrm{V}_{\\mathrm{m}}}{\\sqrt{2}}=\\frac{2}{\\sqrt{2}}=\\sqrt{2}\\) = 1.414 volts<\/li>\n
                    3. Vavg = \\(\\frac{2 \\mathrm{~V}_{\\mathrm{m}}}{\\pi}=\\frac{2 \\times 2}{3.14}\\) = 1.27 volt<\/li>\n
                    4. f = \\(\\frac{1}{\\mathrm{~T}}=\\frac{1}{10 \\times 10^{-3}}=\\frac{1000}{10}\\) = 100Hz<\/li>\n<\/ol>\n

                      Question 9.
                      \nWhat are non-sinusoidal AC waveforms? Where are they used? Draw square wave and sawtooth wave.
                      \nAnswer:
                      \nAny waveform that is not a sine or cosine wave is called a non-sinusoidal AC waveform.
                      \n\"1st
                      \nNon-sinusoidal AC waveforms are used in scanning and digital circuit.<\/p>\n

                      Question 10.
                      \nWhat is AC? Why is DC not used for powering all household electrical devices?
                      \nAnswer:
                      \nAC is that whose amplitude varies periodically with respect to time and changes the polarity at regular intervals.<\/p>\n

                      If DC is used for long-distance transmission, the energy losses become too large. Hence, AC is stepped up and transmitted to a large distance with minimum energy losses.<\/p>\n

                      Five Mark Questions and Answers<\/span><\/p>\n

                      Question 1.
                      \nMention the properties of charges.
                      \nAnswer:<\/p>\n

                        \n
                      1. Like charges repel and unlike charges attract each other.<\/li>\n
                      2. Charges can neither be created nor be destroyed.<\/li>\n
                      3. Charges take only integral values.<\/li>\n
                      4. Charges on a body are not affected by the motion of the body.<\/li>\n
                      5. Charges always reside only on the outer surface of a conductor only.<\/li>\n<\/ol>\n

                        \"1st<\/p>\n

                        Question 2.
                        \nDerive an expression for the effective resistance of resistors connected in series.
                        \n\"1st
                        \nAnswer:
                        \nConsider two resistors R1<\/sub> and R2<\/sub> connected in series to a cell of emf E. In series combination,
                        \nthe resistors are connected end to end and the same current flows through each of them.
                        \nThe emf of the cell equals voltage drops across each of the resistors
                        \nE = V1<\/sub> + V2<\/sub>= IR1<\/sub> + 1R2<\/sub> {from ohms law}
                        \nE = I(R1<\/sub> + R2<\/sub>) \u2192 (1)<\/p>\n

                        The effective or equivalent resistor is a single resistor that has the same effect as a combination of resistors. If Rs. is the effective resistance, E = IR5<\/sub> – (2)
                        \nFrom the equations (1) and (2), IRs<\/sub> = I(R1<\/sub> + R2<\/sub>)
                        \n\u2234 Rs<\/sub> = R1<\/sub> + R2<\/sub>
                        \n\u2234 The effective resistance of resistors in series in the sum of the individual resistances.<\/p>\n

                        Question 3.
                        \nDerive an expression for effective resistance of resistors connected in parallel.
                        \n\"1st
                        \nAnswer:
                        \nConsider two resistors R1<\/sub> and R2<\/sub> connected in parallel between the same two common points. In parallel combination, the voltage across each resistor is the same.<\/p>\n

                        Let I1<\/sub> and I2<\/sub> be the currents through the resistors R1<\/sub> and R2<\/sub>.
                        \nThe current from the cell equals the sum of the currents through the two resistors, i.e.,
                        \nI = I1<\/sub> + I2<\/sub> = \\(\\frac{V}{R_{1}}+\\frac{V}{R_{2}}\\) = V\\(\\left(\\frac{1}{R_{1}}+\\frac{1}{R_{2}}\\right)\\) \u2192 (1)
                        \nEquivalent or effective resistor Rp is a single resistor that has the same effect as a tile parallel combination. If Rp is the effective resistor, then I = \\(\\frac{\\mathrm{V}}{\\mathrm{Rp}}\\) \u2192 (2)<\/p>\n

                        From equations (1) and (2)
                        \n\\(\\frac{\\mathrm{V}}{\\mathrm{Rp}}\\) = V\\(\\left(\\frac{1}{R_{1}}+\\frac{1}{R_{2}}\\right)\\)
                        \n\\(\\frac{1}{\\mathrm{Rp}}=\\frac{1}{\\mathrm{R}_{1}}+\\frac{1}{\\mathrm{R}_{2}}\\)
                        \nHence, the reciprocal of effective resistance of a number of resistors in parallel is the sum of the reciprocals of the individual resistances.<\/p>\n

                        \"1st<\/p>\n

                        Question 4.
                        \nDefine the following terms (i) Frequency (ii) Time period, (iii) Peak value (iv) RMS value, and (v) peak to the peak value of an ac signal.
                        \nAnswer:<\/p>\n

                          \n
                        1. The frequency of AC is the number of AC cycles completed in one second.<\/li>\n
                        2. The time period of AC is the time taken to complete one cycle.<\/li>\n
                        3. The peak value of AC is the maximum voltage or current of the ac in its one of half cycles.<\/li>\n
                        4. RMS value of ac is the equivalent steady DC value which gives the same heating effect as the same rate in a given resistor.<\/li>\n
                        5. The peak to the peak value of AC is the maximum value attained in both positive and negative half cycles of AC.<\/li>\n<\/ol>\n

                          Question 5.
                          \nState and explain venin’s theorem with an example, or Write a procedure to Venice a given circuit with an example.
                          \nAnswer:
                          \nVenin\u2018s theorem states that any linear bilateral network having several voltage sources and resistors can be replaced by a simple circuit having one voltage source VTh<\/sub> in series with one single resistor RTh<\/sub>.<\/p>\n

                          Step 1: Identity and remove load resistor RL<\/sub> to convert circuit as a two-terminal network. Mark the two terminals as A and B.<\/p>\n

                          Step 2: To find VTh<\/sub>, calculate the open-circuit voltage between terminals A and B.
                          \n\"1st
                          \nVTh<\/sub> = VR2<\/sub> = \\(\\frac{V}{R_{1}+R_{2}}\\) \u00d7 R2<\/sub><\/p>\n

                          Step 3: To find RTh, replace all voltage sources in the network with their internal resistances and then find the effective resistance between terminals A and B.
                          \n\"1st
                          \nStep 4: Construct Thevenin’s equivalent circuit having VTh in series with RJh and reconnect the load resistor which was earlier removed.
                          \nIL<\/sub> = \\(\\frac{\\mathrm{V}_{\\mathrm{Th}}}{\\mathrm{R}_{\\mathrm{Th}}+\\mathrm{R}_{\\mathrm{L}}}\\)
                          \nVL<\/sub> = IL<\/sub>RL<\/sub>
                          \n\"1st<\/p>\n

                          \"1st<\/p>\n

                          Question 6.
                          \nState and explain applying superposition theorem to the given circuit.
                          \nAnswer:
                          \nIn any linear bilateral network consisting of two or more independent sources, the resultant current or voltage in any branch is the algebraic sum of the currents or voltages caused by each independent source acting alone, with all other independent sources being replaced by their internal resistances.<\/p>\n

                          Procedure for using Superposition Theorem:
                          \nStep 1: Retain the source V, and replace all other sources with their internal resistances.
                          \n\"1st
                          \n\"1st
                          \nStep 2: Current in various branches and their voltage drops due to V, alone are determined as follows.
                          \n\"1st
                          \nVoltage across R3<\/sub> = V3<\/sub>1<\/sup> = I3<\/sub>1<\/sup> = I3<\/sub>1<\/sup>1<\/sup> = I1<\/sub>1<\/sup>1<\/sup> = Im<\/sub>2<\/sup> \u00d7 \\(\\frac{R_{1}}{R_{1}+R_{3}}\\)<\/p>\n

                          Step 3: Consider V2<\/sub> alone and find the currents and voltages as follows :
                          \n\"1st
                          \n\"1st
                          \nStep 4: Find the algebraic sum of currents and voltage drops in a branch.
                          \nI3<\/sub> = I3<\/sub>1<\/sup> + I3<\/sub>1<\/sup>1<\/sup>; V3<\/sub> = I3<\/sub>1<\/sup> R3<\/sub>, V3<\/sub>1<\/sup>1<\/sup> = I3<\/sub>1<\/sup>1<\/sup> R3<\/sub>
                          \nV3<\/sub> = V3<\/sub>1<\/sup> + V3<\/sub>1<\/sup>1<\/sup><\/p>\n

                          \"1st<\/p>\n

                          Question 7.
                          \nAccording to the maximum transfer theorem, what should be the value of load resistor RL<\/sub> to get maximum power from the 16V battery as shown in the figure below. What is the value of power across RL<\/sub>?
                          \nAnswer:
                          \nTo get maximum power, RL<\/sub> should be equal to Thevenin equivalent resistance.
                          \n\"1st
                          \n\"1st
                          \nRTh<\/sub> = R2<\/sub> + \\(\\frac{\\mathrm{R}_{1} \\mathrm{R}_{3}}{\\mathrm{R}_{1}+\\mathrm{R}_{3}}\\)
                          \nRTh<\/sub> = 2 + \\(\\frac{6 \\times 3}{6 \\times 3}\\) = 2 + \\(\\frac{18}{9}\\) = 2 + 2 = 4\u2126
                          \n\"1st
                          \n\"1st
                          \n\"1st<\/p>\n

                          Question 8.
                          \nFind the branch currents.
                          \n\"1st
                          \nAnswer:
                          \nApply KVL to the mesh ABEFA,
                          \nI1<\/sub>R1<\/sub> + 8(I1<\/sub> + I2<\/sub>) = 12
                          \n4I1<\/sub> + 8 (I1<\/sub> + I2<\/sub>) = 12
                          \n12I1<\/sub> + 8I2<\/sub> = 12 (\u00f74)
                          \n3I1<\/sub> + 2I2<\/sub> = 3 = \u2192 (1)<\/p>\n

                          \"1st<\/p>\n

                          Apply KVL to the loop BCDEB,
                          \n-R2<\/sub>I2<\/sub> – R3<\/sub>I2<\/sub> – R3<\/sub>(I1<\/sub> + I2<\/sub>) = 0
                          \n-5I2<\/sub> – 3I2<\/sub> – 8I1<\/sub> – 8I2<\/sub> = 0
                          \n-8I1<\/sub> – 16I2<\/sub> = 0(\u00f7 – 8)
                          \nI1<\/sub> + 2I2<\/sub> = 0 – (2) .
                          \n3I1<\/sub> + 2I2<\/sub> = 3 – (1)
                          \n-2I1<\/sub> = -3
                          \nI1<\/sub> + I2<\/sub> = 1.5 + (-0.75)
                          \nI1<\/sub> = 1.5 A
                          \nI1<\/sub> + I2<\/sub> = 0.75A
                          \nI2<\/sub> = -0.75 A<\/p>\n

                          Question 9.
                          \nFind the load current and load voltage applying Thevenin\u2019s Theorem.
                          \nAnswer:
                          \n\"1st
                          \nTo find RTh<\/sub>
                          \n\"1st
                          \nRTh<\/sub> = R2<\/sub> + \\(\\frac{\\mathbf{R}_{1} \\mathbf{R}_{3}}{\\mathbf{R}_{1}+\\mathrm{R}_{3}}\\) = 6 + \\(\\frac{6 \\times 3}{6+3}\\) = 6 + 2 = 8<\/p>\n

                          \"1st<\/p>\n

                          To find VTh
                          \n\"1st
                          \nI = \\(\\frac{V}{R_{1}+R_{3}}=\\frac{12}{6+3}\\) = 1.33 A
                          \nVTh<\/sub> = VAB<\/sub> = IRB<\/sub>= 1.33 \u00d7 3 = 3.99 volt.
                          \n\"1st
                          \nIL<\/sub> = \\(\\) = 1.09 A
                          \nVL<\/sub> = IL<\/sub>RL<\/sub> = 109<\/sup> \u00d7 3 = 3.27 volt<\/p>\n

                          Question 10.
                          \nFind the current through 4\u2126 resistors using superposition theorem.
                          \n\"1st
                          \nAnswer:
                          \nStep 1: E1<\/sub> alone ON and E2<\/sub> replaced by a short.
                          \n\"1st
                          \n\"1st
                          \nStep 2: E2<\/sub> alone is ON and E1<\/sub> replaced by short
                          \n\"1st
                          \nRT2<\/sub><\/sub> = R3<\/sub> + \\(\\frac{\\mathrm{R}_{1} \\mathrm{R}_{2}}{\\mathrm{R}_{1}+\\mathrm{R}_{2}}\\) = 4 + \\(\\frac{6 \\times 12}{6+12}\\) = 8\u2126
                          \nIm<\/sub>11<\/sup> = \\(\\frac{E_{2}}{R_{T_{2}}}=\\frac{16}{8}\\) = 2A
                          \nFrom the principle of superposition, current through 4Q resistor is I = Im<\/sub>1<\/sup> – Im<\/sub>1<\/sup>1<\/sup>
                          \n= 3.75 – 2 = 1.75A<\/p>\n

                          1st PUC Electronics Question Bank with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

                          Karnataka 1st PUC Electronics Question Bank Chapter 2 Principles of Electricity, Network Theorems and AC Principles One Mark Questions and Answers Question 1. What is the unit of electric charge? Answer: coulomb (C). Question 2. What is the magnitude of the charge of an electron? Answer: 1.6 \u00d7 10-19 C Question 3. Define potential differences. …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/88292"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=88292"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/88292\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=88292"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=88292"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=88292"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}