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Question
A cell of emf E and internal resistance r is connected to two external resistance R1 and R2 and a perfect ammeter. The current in the circuit is measured in four different situations:
(i) without any external resistance in the circuit
(ii) with resistance R1 only
(iii) with R1 and R2 in series combination
(iv) with R1 and R2 in parallel combination
The currents measured in the four cases are 0.42 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in the order. Identify the currents corresponding to the four cases mentioned above.
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Solution
The current relating to corresponding situations are as follows:
(i) Without any external resistance in the circuit −
`I_1 = E/r`
The current in this case would be maximum, so I1 = 4.2A
(ii) With resistance R1 only −
`I_2 = E/(r + R_1)`
The current in this case will be the second smallest value, so I2 = 1.05 A
(iii) With R1 and R2 in series combination −
`I_3 = E/(r+ (R_1+R_2))`
The current in this case will be minimum as the resistance will be maximum, so I = 0.42 A.
(iv) With R1 and R2 in parallel combination-
`I_4 = E/(r +((R_1R_2)/(R_1 +R_1)))`
The current in this case would be the second largest value, so I = 1.4 A.
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