Question

In the circuit $R_1=R_2$.The value of $E$ and $R_1$ are ________($E-EMF,R_1$-resistance)

• A. $180V,60\Omega$
• B. $120V,60\Omega$
• C. $180V,10\Omega$
• D. $120V,10\Omega$

Answer 1

Answer: (A)

Applying Kirchoff's 2 nd law in loop $ABEFA$
$-IR-(I-1.5)R+E=0$

or $E=IR+(I-1.5)R$
$\Rightarrow E=R[2/-1.5]...(i)$
Now apply KVL in loop $ABCDEFA$, $-IR-1.5R^{\prime }+E=0$
$\Rightarrow E=IR+1.5R^{\prime }...(ii)$
Eliminating / from Equ. (i) and (ii),
$\Rightarrow \frac{1}{2}\left[\frac{E}{R}+1.5\right]=1$[from Eq.(i)]
and $\left[\frac{E-1.5R^{\prime }}{R}\right]=1$
[from Eq. (ii)]
So, $\frac{E}{R}+1.5=\frac{2}{R}\left[E-1.5R^{\prime }\right]$
So, if $R^{\prime }$ were known, then from options we could get the answer.