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**(a) **Three resistors of resistances R1, R2 and R3 are connected **(i)** in series, and**(ii) **in parallel. Write expressions for the equivalent resistance of the combination in each case.**(b)** Two identical resistors of 12 $\Omega$ each are connected to a battery of 3 V. Calculate the ratio of the power consumed by the resulting combinations with minimum resistance and maximum resistance.

(a) (i) __Resistors are connected in series:
__

We know that in a series circuit, the current that flows through each of the resistors is the same, and the voltage across the circuit is the sum of the individual voltage drops across each component.

$\therefore V=V_1+V_2+V_3$

Using Ohm's law, we have-

$V=I\times R$

Putting the value of $V$ for each resistance, we get-

$IR_S=IR_1+IR_2+IR_3$

$IR_S=I(R_1+R_2+R_3)$

$R_S=R_1+R_2+R_3$

Here, $R_S=Equivalent\ resistnce\ in\ series.$

(a) (ii) __Resistors are connected in parallel:__

We know that in a parallel circuit, the voltage across each of the resistors is the same, and the total current is the sum of the currents flowing through each component.

$\therefore I=I_1+I_2+I_3$

Using Ohm's law, we have-

$I=\frac {V}{R}$

Putting the value of $I$ for each resistance, we get-

$\frac {V}{R_P}=\frac {V}{R_1}+\frac {V}{R_2}+\frac {V}{R_3}$

$\frac {1}{R_P}=\frac {1}{R_1}+\frac {1}{R_2}+\frac {1}{R_3}$

Here, $R_P=Equivalent\ resistnce\ in\ parallel.$

(b) Given:

Resistance, $R_1=R_2=12\Omega$

Voltage, $V=3V$.

To find: The ratio of the power consumed by the resulting combinations with minimum resistance (parallel) and maximum resistance (series),$\frac {P_{min}}{P_{max}}$.

Solution: (i) $Maximum\ Resistance \Rightarrow Series\ Combination$

In series, total resistance is given as-

$R_S=R_1+R_2$

$R_S=12+12$

$R_S=24\Omega$

Thus, the total resistance in series, $R_S$ is $24\Omega$

Now,

We know that, power consumed by a resistor is given as-

$Power=\frac{Voltage^2}{Resistance}$ Or, $P = \frac{V^2}{R}$

Substituting the value of $V$ and $R$ we get-

$P_S=\frac{3^2}{24}$

$P_S=\frac{9}{24}$

$P_S=\frac{3}{8}$

$P_S=0.37W$

Thus, the power consumed in series, $P_S$ is 0.37 Watt.

(ii) $Minimum\ Resistance \Rightarrow Parallel\ Combination$

In parallel, total resistance is given as-

$\frac{1}{R_P}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R_P}=\frac{1}{12}+\frac{1}{12}$

$\frac{1}{R_P}=\frac{1+1}{12}$

$\frac{1}{R_P}=\frac{2}{12}$

$\frac{1}{R_P}=\frac{1}{6}$

$R_P=6\Omega$

Thus, the effective resistance in parallel, $R_P$ is $6\Omega$

Now,

We know that, power consumed by a resistor is given as-

$Power=\frac{Voltage^2}{Resistance}$ Or, $P = \frac{V^2}{R}$

Substituting the value of $V$ and $R$ we get-

$P_R=\frac{3^2}{6}$

$P_R=\frac{9}{6}$

$P_R=\frac{3}{2}$

$P_R=1.5W$

Thus, the power consumed in parallel, $P_R$ is 1.5 Watt.

Ratio of power consumed in the two combinations is given as-

$\frac {P_{min}}{P_{max}}=\frac {P_S}{P_R}=\frac{\frac {3}{8}}{\frac {3}{2}}=\frac {3}{8}\times {\frac {2}{3}}=\frac {2}{8}=\frac {1}{4}=1:4$

Thus, the ratio of the power consumed by the resulting combinations with minimum resistance (parallel) and maximum resistance (series), is** 1:4.**