In the circuit cell are of equal `emf E` but of difference internal resistance `r_(1) = 6 Omega` reading of the ideal voltmeter connected across cell `1` is zero The value of the external resistance `R` is ohm is equal to

Options

A. `10 Omega`
B. `4 Omega`
C. `2 Omega`
D. `2.4 Omega`

Answer

c

Explanation

Current in the circuit
`1 = (E+E)/(r_(1) + (r_(2) + R)) = (2E)/(6+4+R) = (2E)/(10+R)`
Potential difference across `(R+ r_(2))`
`= (2E)/((10+R)) (R+r_(2))`
As voltmeter connecting cell `1` records zero reading therefore the e.m.f of cell `2` is equal to voltage drop across `(R+r_(2))` hence
`E = (2E)/((10+R))(R+r_(2))= (2E(R+4))/(10+R)`
On solving `R = 2 Omega`