Question

Find the current through the resistance $R$ in figure if
(a) $R=12$ (b) $R=48\Omega$

Answer 1

Diode from the circuit can be removed as the battery with higher potential is make the diode reverse biased.

So for$R=12\Omega$

$-4V-6V+12\Omega \times I+12\Omega \times I=0$

$I=\frac{10}{24}A$

So for$R=48\Omega$

$-4V-6V+48\Omega \times I+12\Omega \times I=0$

$I=\frac{10}{60}A$