In the following circuit, $5$ $\Omega$ resistor develops $45$ $J/s$ due to current flowing through it. The power developed per second across $12$ $\Omega$ resistor is ............. $W$
A$16$
B$192$
C$36$
D$64$
Diffcult
Download our app for free and get started
B$192$
b (b) $\frac{{{i_1}}}{{{i_2}}} = \frac{{15}}{5} = \frac{3}{1}$… $(i)$
Also $\frac{H}{t} = {i^2}R \Rightarrow 45 = {({i_1})^2} \times 5$
$ \Rightarrow $ ${i_1} = 3\,A$ and from equation $(i)$ ${i_2} = 1\,A$
So $i = {i_1} + {i_2} = 4\,A$
Hence power developed in $12\,\Omega $ resistance $P = {i^2}R = {(4)^2} \times 12 = 192\,W$
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
The current in a simple series circuit is $5.0\, amp$. When an additional resistance of $2.0\, ohms$ is inserted, the current drops to $4.0\, amp$. The original resistance of the circuit in $ohms$ was
A $100\, W$ bulb $B_1$, and two $60\,W$ bulbs $B_2$ and $B_3$, are connected to a $250\, V$ source, as shown in the figure. Now $ W_1, W_2$ and $W_3$ are the output powers of the bulbs $B_1, B_2$ and $B_3$, respectively. Then
$A$ brass disc and a carbon disc of same radius are assembled alternatively to make a cylindrical conductor. The resistance of the cylinder is independent of the temperature. The ratio of thickness of the brass disc to that of the carbon disc is [$\alpha$ is temperature coefficient of resistance and Neglect linear expansion ]