Question

All bulbs consume same power. The resistance of bulb 1 is $36\Omega$. What is the voltage output of the battery if the power of each bulb is 4W?

• A. 12 V
• B. 16 V
• C. 24 V
• D. None of these

Answer 1

The correct option is B 16 V

As bulbs 2 and 3 are in series so current through both is same. i.e, $i_2=i_3=I_b$.

As 1 and combine (2,3) are in parallel so, $V_1=(V_2+V_3)=V$ (say)

As all consume same power so, $P_1=P_2=P_3=P_4$

Now, $P_2=P_{3}or{I}_b^2{R}_2=I_b^2{R}_3\Rightarrow R_2=R_3$

Since, $R_2=R_3$ so $V_2=V_3=V/2$

Now, $P_1=\frac{V_1^2}{R_1}=\frac{V^2}{36}$

and $P_3=\frac{V_3^2}{R_3}=\frac{(V/2{)}^2}{R_3}=\frac{V^2}{4R_3}$

Since, $P_1=P_3\Rightarrow \frac{V^2}{36}=\frac{V^2}{4R_3}\to R_3=9\Omega$

Also, $R_2=R_3=9\Omega$

Now current, $I_a=\frac{(R_2+R_3)}{R_1+(R_2+R_3)}I=\frac{18}{54}I=I/3$ and $I_4=I$

Since, $P_1=P_4\Rightarrow I_a^2{R}_1=I_4^2{R}_4$

$\Rightarrow (I/3{)}^2\times 36=I^2{R}_4\Rightarrow R_4=4\Omega$

Given: $P_4=4W$ so $4=I^2{R}_{4}or4=I^2\left(4\right)$

Thus, $I=1A$

Equivalent resistance of the circuit is: $R_{eq}=R_4+\frac{R_1(R_2+R_2)}{R_1+R_2+R_3}=4+\frac{36(9+9)}{36+9+9}=4+12=16\Omega$

Voltage of source is: $\epsilon =I{R}_{eq}=16V$