The total power dissipated in watts in the ci | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) First,calculate the equivalent resistance of the parallel combination of $6 \Omega$ and $3 \Omega$ resistors:$\frac{1}{R_p} = \frac{1}{6} + \frac{

Vedclass · Accepted Answer

$54$

Vedclass · Answer

(D) First,calculate the equivalent resistance of the parallel combination of $6 \Omega$ and $3 \Omega$ resistors:$\frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2} \implies R_p = 2 \Omega$.Now,this $R_p$ is in series with the $4 \Omega$ resistor.The total equivalent resistance of the circuit is $R_{eq} = R_p + 4 \Omega = 2 \Omega + 4 \Omega = 6 \Omega$.The total power dissipated in the circuit is given by the formula $P = \frac{V^2}{R_{eq}}$,where $V = 18 \ V$.$P = \frac{18^2}{6} = \frac{324}{6} = 54 \ W$.Therefore,the correct option is $D$.

The total power dissipated in watts in the circuit shown is ............. $W$.

Similar Questions

What is the equivalent resistance across $A$ and $B$ in the figure shown,if $R = 3\,\Omega$?

The resistance between points $A$ and $C$ in the given network is

Two resistances are joined in parallel whose resultant is $6/8 \, \Omega$. One of the resistance wires is broken and the effective resistance becomes $2 \, \Omega$. Then the resistance in ohm of the wire that got broken was:

Two resistors are joined in parallel,and their resultant resistance is $\frac{6}{5} \,\Omega$. One of the resistance wires is broken,and the effective resistance becomes $2 \,\Omega$. Then the resistance (in $\Omega$) of the wire that got broken is ..........

The equivalent resistance between $A$ and $B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module