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A resistance of 9 ohm is connected to the terminals of a cell. A voltmeter connected across the cell reads 1.8V . When a resistance of 10 ohm is connected in series with 9 ohm the voltmeter reading changes to 1.9V . Calculate the emf of the cell and its internal resistance |
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Answer» r,● Answer-r = 1 ohmE = 2 V● Explanation-R1 = 9 ohmR2 = 10 ohmV1 = 1.8 VV2 = 1.9 V# Solution-Let E be emf of cell and r be internal RESISTANCE.In first situation,I1 = V1 / R1I1 = 1.8 / 9I1 = 0.2 ABut, we knowI1 = E / (R1+r) 0.2 = E / (9+r) ...(1)In second situation,R' = R1 + R2 R' = 9 + 10R' = 19 ohmIn this situation,I2 = V2 / R'I2 = 1.9 / 19I2 = 0.1 ABut, we knowI2 = E / (R'+r) 0.1 = E / (19+r) ...(2)Dividing (1) by (2),2 = (19+r) / (9+r)18 + 2r = 19 + RR = 1 ohmTherefore,E = I1 (R1+r)E = 0.2 (9+1)E = 2 VHence, emf of cell is 2 V and internal resistance is 1 ohm.Hope this helps... |
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