Question

All bulbs consume same power. The resistance of bulb 1 is $36\Omega$. What is the resistance of bulb-4?

• A. $4\Omega$
• B. $9\Omega$
• C. $12\Omega$
• D. $18\Omega$

Answer 1

The correct option is A $4\Omega$

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

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

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