NCERT Solutions

We need to calculate the current flowing through every individual resistor branch in the provided bridge circuit.

Concept: Kirchhoff's Circuit Laws

The solution utilizes Kirchhoff's Circuit Laws. Kirchhoff's Current Law (KCL) states that the sum of currents entering a junction equals the sum of currents leaving it. Kirchhoff's Voltage Law (KVL) states that the algebraic sum of potential differences around any closed loop is zero ($\sum V = 0$).

Given:

• Resistance $R_{AB} = 10\,\Omega$, $R_{BC} = 5\,\Omega$
• Resistance $R_{AD} = 5\,\Omega$, $R_{DC} = 10\,\Omega$
• Galvanometer/Bridge resistance $R_{BD} = 5\,\Omega$
• External resistance $R_{ext} = 10\,\Omega$
• Battery voltage $V = 10\,\text{V}$

Solving:

1. Define Branch Currents:

   Let the current from the battery be $I$. At node $A$, it splits into $I_1$ (through $AB$) and $I_2$ (through $AD$).

   • Current in $AB = I_1$
   • Current in $AD = I_2$
   • Current in $BD = I_3$ (from $B$ to $D$)
   • Current in $BC = I_1 - I_3$ (by KCL at $B$)
   • Current in $DC = I_2 + I_3$ (by KCL at $D$)
   • Total current $I = I_1 + I_2$

2. Apply KVL to Loops:

   • Loop ABDA:

   $$\begin{aligned} 10 I_1 + 5 I_3 - 5 I_2 &= 0 \\ 2 I_1 + I_3 - I_2 &= 0 \quad \dots (1) \end{aligned}$$

   • Loop BCDB:

   $$\begin{aligned} 5(I_1 - I_3) - 10(I_2 + I_3) - 5 I_3 &= 0 \\ 5 I_1 - 5 I_3 - 10 I_2 - 10 I_3 - 5 I_3 &= 0 \\ 5 I_1 - 10 I_2 - 20 I_3 &= 0 \\ I_1 - 2 I_2 - 4 I_3 &= 0 \quad \dots (2) \end{aligned}$$

   • Loop ADCA (including battery and external resistor):

   $$\begin{aligned} 5 I_2 + 10(I_2 + I_3) + 10(I_1 + I_2) &= 10 \\ 10 I_1 + 25 I_2 + 10 I_3 &= 10 \\ 2 I_1 + 5 I_2 + 2 I_3 &= 2 \quad \dots (3) \end{aligned}$$

3. Solve the Linear Equations:

   From (1), $I_2 = 2 I_1 + I_3$. Substitute this into (2):

   $$\begin{aligned} I_1 - 2(2 I_1 + I_3) - 4 I_3 &= 0 \\ I_1 - 4 I_1 - 2 I_3 - 4 I_3 &= 0 \\ -3 I_1 - 6 I_3 &= 0 \\ I_1 &= -2 I_3 \end{aligned}$$

   Substitute $I_1 = -2 I_3$ into $I_2 = 2 I_1 + I_3$:

   $$I_2 = 2(-2 I_3) + I_3 = -3 I_3$$

   Now substitute $I_1$ and $I_2$ into (3):

   $$\begin{aligned} 2(-2 I_3) + 5(-3 I_3) + 2 I_3 &= 2 \\ -4 I_3 - 15 I_3 + 2 I_3 &= 2 \\ -17 I_3 &= 2 \\ I_3 &= -\frac{2}{17}\,\text{A} \end{aligned}$$

   Calculating other values:

   • $I_1 = -2\left(-\frac{2}{17}\right) = \frac{4}{17}\,\text{A}$
   • $I_2 = -3\left(-\frac{2}{17}\right) = \frac{6}{17}\,\text{A}$

4. Calculate Final Branch Currents:

   • $I_{AB} = I_1 = \frac{4}{17}\,\text{A}$
   • $I_{AD} = I_2 = \frac{6}{17}\,\text{A}$
   • $I_{BD} = I_3 = -\frac{2}{17}\,\text{A}$ (Current flows from $D$ to $B$)
   • $I_{BC} = I_1 - I_3 = \frac{4}{17} - \left(-\frac{2}{17}\right) = \frac{6}{17}\,\text{A}$
   • $I_{DC} = I_2 + I_3 = \frac{6}{17} + \left(-\frac{2}{17}\right) = \frac{4}{17}\,\text{A}$
   • $I_{total} = I_1 + I_2 = \frac{10}{17}\,\text{A}$

The currents in the branches are:

• Current in $AB = \frac{4}{17}\,\text{A}$
• Current in $BC = \frac{6}{17}\,\text{A}$
• Current in $AD = \frac{6}{17}\,\text{A}$
• Current in $DC = \frac{4}{17}\,\text{A}$
• Current in $BD = -\frac{2}{17}\,\text{A}$ (or $\frac{2}{17}\,\text{A}$ from $D$ to $B$)
• Total current $I = \frac{10}{17}\,\text{A}$