Question

In the circuit shown in figure, match the following.

Table-1 Table-2

$\left(\text{A}\right)$ Minimum current will flow through.	$\left(\text{P}\right)2\Omega$
$\left(\text{B}\right)$ Maximum current will flow through.	$\left(\text{Q}\right)4\Omega$
$\left(\text{C}\right)$ Maximum power will be genrated across.	$\left(\text{R}\right)3\Omega$
$\left(\text{A}\right)$ Minimum power will be generated across.	$\left(\text{S}\right)5\Omega$

• A. $\text{A}\to \text{Q},\text{B}\to \text{P},\text{C}\to \text{R},\text{D}\to \text{Q}$
• B. $\text{A}\to \text{P},\text{B}\to \text{Q},\text{C}\to \text{R},\text{D}\to \text{S}$
• C. $\text{A}\to \text{Q},\text{B}\to \text{P},\text{C}\to \text{p},\text{D}\to \text{R}$
• D. $\text{A}\to \text{S},\text{B}\to \text{P},\text{C}\to \text{R},\text{D}\to \text{P}$

Answer 1

The correct option is A $\text{A}\to \text{Q},\text{B}\to \text{P},\text{C}\to \text{R},\text{D}\to \text{Q}$

In series :

Current passing through resistors connected in series will remain same , where as potential drop across the resistor depends on the value of resistance.

In parallel :

Potential difference across the resistors remains same where as the current distributes in the inverse ratio of resistances.

Let $R$ be the effective resistance and $I$ be the total current in the circuit.

Current passing through $2\Omega$ resistor , $I_2=\frac{I{R}_4}{R_2+R_4}=\frac{4I}{6}=\frac{2I}{3}$

Current passing through $4\Omega$ resistor , $I_4=\frac{I{R}_2}{R_2+R_4}=\frac{2I}{6}=\frac{I}{3}$

Current passing through $3\Omega$ resistor , $I_3=\frac{I{R}_5}{R_3+R_5}=\frac{5I}{8}$

Current passing through $5\Omega$ resistor , $I_5=\frac{I{R}_3}{R_3+R_5}=\frac{3I}{8}$

Therefore, $I_4<I_5<I_3<I_2.......\left(1\right)$

So, $\text{A}\to \text{Q}$ and $\text{B}\to \text{P}$

Power generated across $2\Omega$ resistor is , $P_2=I_2^2{R}_2=\frac{8I^2}{9}$

Power generated across $4\Omega$ resistor is , $P_4=I_4^2{R}_4=\frac{4I^2}{9}$

Power generated across $3\Omega$ resistor is , $P_3=I_3^2{R}_3=\frac{75I^2}{64}$

Power generated across $5\Omega$ resistor is , $P_5=I_5^2{R}_5=\frac{45I^2}{64}$

Therefore, $P_3>P_5>P_2>P_4....\left(2\right)$

So, $\text{C}\to \text{R}$ and $\text{D}\to \text{Q}$

Hence, option (a) is the correct answer.