In the following circuit, 5Ωresistor develops 45 J/sdue to current flowing through it. The power developed across 12Ω resistor is

$R_{e f f} = 12 + \frac{15 \times 5}{20}$ $= \frac{63}{4}$

$V = i \times \frac{63}{4}$

$i = \frac{4 V}{63}$

$i = i_{1} + i_{2}$

$15 i_{2} = 5 i_{1}$

$i_{1} = 3 i_{2} \Rightarrow i_{2} = \frac{i_{1}}{3}$

$i = i_{1} + \frac{i_{1}}{3} = \frac{4 i_{1}}{3}$

$P = i_{1}^{2} \times 5$

$45 = i_{1}^{2} \times 5 \Rightarrow i_{1}^{2} = 9$

$i_{1} = 3$

$∴ i = 4$

power developed through $12 Ω$

Resistor is $= P = i^{2} \times R$

$= 4^{2} \times 12$

$= 16 \times 12 = 192 W$

In the following circuit, $5 Ω$ resistor develops $45 J / s$ due to current flowing through it. The power developed across $12 Ω$ resistor is