Question

What is the magnitude of $V_0$, when power loss in 10 $\Omega$ resistor is 90 watt?

• A. 15 V
• B. 9 V
• C. 6 V
• D. 3 V

Answer 1

The correct option is D 3 V

From the first circuit
$V_0=i_1[1+\frac{3\times 6}{3+6}]=3i_1$

From the second circuit,
Power across $10\Omega =90W$

$i_L^2\times 10=90$

$i_L=\sqrt{9}=3A.$

$6V_a=2i_L=2\times 3=6A$

Hence $6V_a=6$

$V_a=1V$

$Fromthefigure,V_a=i_1\times 1=i_1$

So, $i_1=1A$

From equation (i), $V_0=3i_1=3\times 1=3V$