InterviewSolution
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InterviewSolution
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Two cells of emf 1.5 V and 2 V and internal resistance 1Omega and 2Omegarespectively are connected inparallel to pass a current in the same direction through an external resistance of 5Omega.(i) Draw the circuit diagram.(ii) Using Kirchhoff's laws, calculate the current through each branch of the circuit and potential difference across the 5Omegaresistor. |
Answer» Solution :(i) The circuit diagram has been SHOWN in Fig. (ii) In mesh ACRDBA, applying Kirchhoff.s second LAW, we have ` - (I_1 + I_2) .5- I_1 .1 + 1.5 = 0` `rArr6I_1 + 5I_2 = 1.5 `....(i) and for mesh CRDC, we have ` - (I_1 + I_2) .5 - I_2 .2 + 2= 0` ` RARR 5I_1 + 7I_2 = 2`....(ii) On solving (i) and (ii), we get `I_1 = 1/34 A` and `I_2 = 9/34 A` Current through external `(5OMEGA)` resistor `= I_1 + I_2 = 1/34 + 9/34= 10/34 A = 5/17 A` ` therefore `Potential difference across the 5`Omega`resistor `V = (I_1 + I_2) R = 5/17 xx 5 = 25/17 V` |
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