MP Board · Class 12 · Physics · Current ElectricityA network of resistors is connected to a 12 V battery with internal resistance 1 Ω. The circuit consists of three resistors: $2\,\Omega$, $3\,\Omega$, and $6\,\Omega$ connected in parallel, and this parallel combination is connected in series with a $4\,\Omega$ resistor. (a) Calculate the equivalent resistance of the network. (b) Calculate the total current flowing in the circuit. (c) Calculate the potential difference across the $4\,\Omega$ resistor and the parallel combination.

Given Data:

• EMF of the battery ($E$) = $12\text{ V}$
• Internal resistance ($r$) = $1,\Omega$
• Resistors in parallel: $R_1 = 2,\Omega$, $R_2 = 3,\Omega$, $R_3 = 6,\Omega$
• Resistor in series: $R_4 = 4,\Omega$

Step (a): Calculate the equivalent resistance of the network\nFirst, find the equivalent resistance of the parallel combination ($R_p$):

$$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$ $$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$$ \nTaking the LCM of denominators (which is 6): $$\frac{1}{R_p} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1,\Omega^{-1}$$ $$R_p = 1,\Omega$$ \nNow, this parallel combination is in series with $R_4 = 4,\Omega$. Therefore, the total external resistance ($R$) of the circuit is: $$R = R_p + R_4 = 1,\Omega + 4,\Omega = 5,\Omega$$

Step (b): Calculate the total current flowing in the circuit\nThe total resistance of the circuit including internal resistance is:

$$R_{\text{total}} = R + r = 5,\Omega + 1,\Omega = 6,\Omega$$ \nUsing Ohm's Law, the total current ($I$) supplied by the battery is: $$I = \frac{E}{R_{\text{total}}} = \frac{12\text{ V}}{6,\Omega} = 2\text{ A}$$

Step (c): Calculate the potential difference across the $4,\Omega$ resistor and the parallel combination

1. Potential difference across the $4,\Omega$ resistor ($V_4$): $$V_4 = I \times R_4 = 2\text{ A} \times 4,\Omega = 8\text{ V}$$

2. Potential difference across the parallel combination ($V_p$): $$V_p = I \times R_p = 2\text{ A} \times 1,\Omega = 2\text{ V}$$ (Alternatively, using Kirchhoff's Voltage Law: $12\text{ V} - 8\text{ V} - (2\text{ A} \times 1,\Omega) = 2\text{ V}$)