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Two cells of emfs E1 and E2 and of negli...

Two cells of emfs `E_1` and `E_2` and of negligible internal resistances are connected with two variable resistors as shown in Fig. A2.4. When the galvanometer shows no deflection, the values of the resistances are P and Q. What is the value of the ratio `E_2/E_1`?
`

A

`P/Q`

B

`P/(P+Q)`

C

`Q/(P+Q)`

D

`(P+Q)/P`

Text Solution

Verified by Experts

The correct Answer is:
B
b. Potential difference `V_P` across P as determined from `E_1` id given
by `V_P = (E_1/(P+Q))P.`
Potential `V_P` across P as determined from `E_2` is same as `E_2`
because no current is drawn, i.e., `V_P = E_2`
Therefore,
`E_2 = E_1(P/(P+Q)) or E_2//E_1 = (P/(P+Q))` .
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