The bulb which glows with maximum intensity i | vedclass

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(A) First,simplify the circuit. The circuit consists of two parts in series: part $A$ and part $B$.Part $A$ has resistors of $2 \Omega, 3 \Omega$,and

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$4 \Omega$ bulb

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(A) First,simplify the circuit. The circuit consists of two parts in series: part $A$ and part $B$.Part $A$ has resistors of $2 \Omega, 3 \Omega$,and $6 \Omega$ in parallel. The equivalent resistance $R_A$ is given by $\frac{1}{R_A} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1 \Omega$,so $R_A = 1 \Omega$.Part $B$ has resistors of $4 \Omega$ and $5 \Omega$ in parallel. The equivalent resistance $R_B$ is given by $\frac{1}{R_B} = \frac{1}{4} + \frac{1}{5} = \frac{5+4}{20} = \frac{9}{20} \Omega$,so $R_B = \frac{20}{9} \Omega \approx 2.22 \Omega$.Since $R_B > R_A$,the potential drop across part $B$ $(V_B)$ is greater than the potential drop across part $A$ $(V_A)$.For resistors in parallel,power $P = \frac{V^2}{R}$. Since $V_B$ is constant for both resistors in part $B$,the resistor with the smaller resistance will have higher power dissipation.Comparing the $4 \Omega$ and $5 \Omega$ resistors in part $B$,the $4 \Omega$ resistor has lower resistance and thus dissipates more power.Therefore,the $4 \Omega$ bulb glows with maximum intensity.

The bulb which glows with maximum intensity in the given circuit is

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