Question

All bulbs consume same power. The resistance of bulb $1$ is $36\Omega$.


$\left(ii\right)$ What is the voltage output of the battery , if the power of each bulb is $4\text{W}?$

• A. $12\text{V}$
• B. $16\text{V}$
• C. $24\text{V}$
• D. $32\text{V}$

Answer 1

The correct option is B $16\text{V}$

From the figure, Bulbs $2$ and $3$ are in series and both consume same power.



Therefore , $R_2=R_3$

Further, $V_1=V_{23}$ and $V_2=V_3$

Thus, $V_2=V_3=\frac{V_1}{2}$

Since, Power $\left(P\right)=\frac{V^2}{R}$

We can say, resistance should becomes \(\left{\dfrac{1}{4}\right)^{th}\).

Hence, $R_2=R_3=\frac{R_1}{4}=9\Omega$

Now, $R_{23}=9+9=18\Omega$ and $R_1=36\Omega$

$\therefore i_{23}=2i_1$, also $i_4=i_{23}+i_1=3i_1$ (Which means current passing through $4$ become $3$ times)

That implies, resistance of $4$ should be ${\left(\frac{1}{9}\right)}^{th}$ of resistance $1$.

$R_4=\frac{R_1}{9}=4\Omega$

For the equivalent resistance, $R_{net}=\frac{R_1\times R_{23}}{R_1+R_{23}}+R_4=\frac{18\times 36}{18+36}+4=16\Omega$

Given, power of each bulb becomes $4\text{W}$

Therefore , $P_{net}=4\times \left(4\text{W}\right)=16\text{W}$

Voltage output of battery , $E=\sqrt{P_{net}{R}_{net}}=16\text{V}$