Two cells P and Q each of emf 2.16 V are connected in series with a resistor of 19.6 `Omega` An ideal voltmeter reads 2V, when connected across the cell P and 1.92 V when connected across the cell Q. The ratio of the internal resistance ot the cell P and Q is

Options

A. `1:2`
B. `2:3`
C. `3:4`
D. `1:3`

Answer

B

Explanation

Emf of each cell = 216 V
Emf of circuit = 4.32 V
Current in circuit,
`1=(4.32)/(r_(1)+r_(2)+19.6)`
Now, using `V = E-1*r ` for both cells , we have
for `p, 2=216-(4.32r_(1))/(r_(1)+r_(2)+19.6)`
and for `Q1.92=216-(4.32r_(2))/(r_(1)+r_(2)+19.6)`
So, we have,
`0.16=(4.32r_(2))/(r_(1)+r_(2)+19.6)`
and,
`0.24=(4.32r_(2))/(r_(1)+r_(2)+19.6)`
from Eqs.(i) and (ii) , we get
`r_(1):r_(2)=2:3`