Question 73: Current Electricity

Question

Refer to the circuit diagram given below. The heat generated across the $6 \Omega$ resistance in 100 second is $\frac{\alpha}{100} \mathrm{~J}$. The value of $\alpha$ is $\_\_\_\_$ . (Nearest integer)

Accepted Answer

The heat $(H)$ generated in a resistor $(R)$ due to a current (i) flowing for a time $(t)$ is :
$$ \mathrm{H}=\mathrm{i}^2 \mathrm{Rt} $$
Let $V_A$ be the potential at the junction $A$. Using Kirchhoff's current law, the sum of currents $=0$ :
$\frac{V_A-3}{6}+\frac{V_A-0}{4}+\frac{V_A-2}{3}=0$
$$\Rightarrow $$ $2\left(V_A-3ight)+3 V_A+4\left(V_A-2ight)=0$
$$\Rightarrow $$ $2 V_A-6+3 V_A+4 V_A-8=0$
$$\Rightarrow $$ $$ 9 V_A=14 \Rightarrow V_A=\frac{14}{9} V $$
So, the current through the $6 \Omega$ resistor is,
$$ \mathrm{i}=\frac{3-\mathrm{V}_{\mathrm{A}}}{6}=\frac{3-14 / 9}{6}=\frac{27-14}{9 	imes 6}=\frac{13}{54} \mathrm{~A} $$
The time is $\mathrm{t}=100 \mathrm{~s}$.
Hence, total heat generated in $6 \Omega$ is;
$$ H=\left(\frac{13}{54}ight)^2 	imes 6 	imes 100=\frac{169}{2916} 	imes 600 \approx 34.77 \mathrm{~J} $$
It is given that $\mathrm{H}=\frac{\alpha}{100}$, then $\alpha=\mathrm{H} 	imes 100$ :
$$ \Rightarrow \alpha \approx 3477 $$
Therefore, the correct answer is 3477.

Current Electricity

Refer to the circuit diagram given below. The heat generated across the 6 \Omega resistance in 100 second is \frac{\alpha}{100} \mathrm{~J}. The value of \alpha is \_\_\_\_ . (Nearest integer)

Integer / Numeric Answer Type