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A battery of emf 10 V and internal resistance 3Omegais connected to a resistor. If the current in thecircuit is 0.5 A, what is the resistance of the resistor ? What is the terminal voltage of the batterywhen the circuit is closed ? |
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Answer» Solution :Here `epsi = 10 V,R= 3Omega`and I = 0.5 A As `I = (epsi)/((R + r))`,hence external resistance R = `epsi/I - r= (10)/(0.5)- 3 = 20 - 3 = 17 OMEGA` and the terminal VOLTAGE of battery `V = epsi-Ir = 10-0.5xx3 = 10-1.5 = 8.5 V` |
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