Question

In the circuit shown in figure, if ammeter and voltmeter are ideal, then the consumed power in $9\Omega$ resistor will be

• A. $3.33W$
• B. $4W$
• C. $1.44W$
• D. $500W$

Answer 1

The correct option is C $1.44W$

$I=\frac{30}{R_{net}}$

$R_{net}=20+5=25$

$I_t=\frac{-30}{25}=\frac{6}{5}A$

voltage at $20\Omega =\frac{6}{5}\times 20=24V$

all parallely connected resistors

$v=6V$

$R=9+6=15\Omega$

$\frac{V}{R}=I$ through $9$ & $6\Omega$ resistor

$\frac{6}{15}=I$

Power consumed by $9\Omega =I^{2}R$

$\frac{6\times 6}{15\times 15}\times 9=1.4W$