Question

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

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

Answer 1

The Correct Option is A

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

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