Question

The current drawn from the battery by the network of resistors as shown in the figure is

• A. 2Ω

Answer 1

Answer: (A)

2A

Answer 2

The circuit diagram shows a network of resistors connected to a 4V battery. We need to find the total current drawn from the battery. The circuit can be analyzed using nodal analysis or by simplifying series and parallel combinations. Let's redraw the circuit and label the nodes. Let the negative terminal of the battery be at node A and the positive terminal be at node C. So, the potential difference between C and A is 4V, i.e., $V_C - V_A = 4V$.

The resistors are connected as follows:

1. A 1Ω resistor between A and B.
2. A 2Ω resistor between A and D, where D is the junction of the 5Ω and 4Ω resistors.
3. A 5Ω resistor between B and D.
4. A 4Ω resistor between D and C.
5. A 2Ω resistor between B and C.

The circuit diagram is a bridge-like structure. Let's calculate the equivalent resistance of the network between A and C. We can use nodal analysis. Let $V_A = 0$ V, then $V_C = 4$ V. Let the potential at node B be $V_B$ and at node D be $V_D$.

Applying Kirchhoff's Current Law at node B:

$\frac{V_A - V_B}{1} + \frac{V_D - V_B}{5} + \frac{V_C - V_B}{2} = 0$

$\frac{0 - V_B}{1} + \frac{V_D - V_B}{5} + \frac{4 - V_B}{2} = 0$

$-V_B + \frac{V_D - V_B}{5} + \frac{4 - V_B}{2} = 0$

Multiplying by 10: $-10V_B + 2(V_D - V_B) + 5(4 - V_B) = 0$

$-10V_B + 2V_D - 2V_B + 20 - 5V_B = 0$

$-17V_B + 2V_D + 20 = 0 \implies 17V_B - 2V_D = 20$ (Equation 1)

Applying Kirchhoff's Current Law at node D:

$\frac{V_A - V_D}{2} + \frac{B - V_D}{5} + \frac{V_C - V_D}{4} = 0$

$\frac{0 - V_D}{2} + \frac{V_B - V_D}{5} + \frac{4 - V_D}{4} = 0$

$-\frac{V_D}{2} + \frac{V_B - V_D}{5} + \frac{4 - V_D}{4} = 0$

Multiplying by 20: $-10V_D + 4(V_B - V_D) + 5(4 - V_D) = 0$

$-10V_D + 4V_B - 4V_D + 20 - 5V_D = 0$

$4V_B - 19V_D + 20 = 0 \implies 4V_B - 19V_D = -20$ (Equation 2)

Now we solve the system of linear equations for $V_B$ and $V_D$.

From Equation 1: $2V_D = 17V_B - 20 \implies V_D = \frac{17V_B - 20}{2}$

Substitute into Equation 2: $4V_B - 19\left(\frac{17V_B - 20}{2}\right) = -20$

$8V_B - 19(17V_B - 20) = -40$

$8V_B - 323V_B + 380 = -40$

$-315V_B = -420$

$V_B = \frac{-420}{-315} = \frac{420}{315} = \frac{4 \times 105}{3 \times 105} = \frac{4}{3}$ V.

Now find $V_D$: $V_D = \frac{17V_B - 20}{2} = \frac{17(4/3) - 20}{2} = \frac{68/3 - 60/3}{2} = \frac{8/3}{2} = \frac{4}{3}$ V.

Since $V_B = V_D = 4/3$ V, the potential difference across the 5Ω resistor is $V_B - V_D = 4/3 - 4/3 = 0$. This means no current flows through the 5Ω resistor. The 5Ω resistor is effectively short-circuited in terms of current flow through it, but it still affects the equivalent resistance of the network. This configuration, where the ratio of resistances in two arms is equal to the ratio of resistances in the other two arms of the bridge (excluding the bridge resistor), results in zero current through the bridge resistor. However, in our case, the ratios are $R_{AB}/R_{AD} = 1/2$ and $R_{BC}/R_{DC} = 2/4 = 1/2$. So, $\frac{R_{AB}}{R_{AD}} = \frac{R_{BC}}{R_{DC}} = \frac{1}{2}$. This is a balanced Wheatstone bridge with the 5Ω resistor as the bridge resistor.

Since the bridge is balanced, the 5Ω resistor can be removed from the circuit without affecting the current distribution in the other resistors or the equivalent resistance between A and C. The circuit simplifies to: A 1Ω resistor in series with a 2Ω resistor, and this combination is in parallel with a 2Ω resistor in series with a 4Ω resistor.

Upper branch resistance: $R_{upper} = R_{AB} + R_{BC} = 1Ω + 2Ω = 3Ω$.

Lower branch resistance: $R_{lower} = R_{AD} + R_{DC} = 2Ω + 4Ω = 6Ω$.

These two branches are connected in parallel between A and C.

Equivalent resistance $R_{eq}$ between A and C is:

$\frac{1}{R_{eq}} = \frac{1}{R_{upper}} + \frac{1}{R_{lower}} = \frac{1}{3Ω} + \frac{1}{6Ω} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$

$R_{eq} = 2Ω$.

The total current drawn from the battery is given by Ohm's law:

$I = \frac{V}{R_{eq}}$

where V is the battery voltage (4V) and $R_{eq}$ is the equivalent resistance (2Ω).

$I = \frac{4V}{2Ω} = 2A$.