Class 11 - Unit # 10 : DC Circuits - Solved Numericals

Physics Numerical Sheet
DC Circuits

1. The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.5 \(\Omega\), what is the maximum current that can be drawn from the battery?

Data:

• Electromotive force (\(E\)) = \(12\text{ V}\)
• Internal resistance (\(r\)) = \(0.5\text{ }\Omega\)
• Maximum current (\(I_{\max}\)) = ?

Solution:

For maximum current output, the external resistance connected to the circuit is considered to be zero. According to Ohm’s Law:

$$I_{\max} = \frac{E}{r}$$

Substituting the given variables:

$$I_{\max} = \frac{12}{0.5} = 24\text{ A}$$

RESULT: The maximum current that can be drawn from the battery is 24 A.

2. A negligibly small current is passed through a wire of length 12 m and uniform cross-section \(4.0 \times 10^{-7}\text{ m}^2\), and its resistance is measured to be 6.0 \(\Omega\). What is the resistivity of the material at the temperature of the experiment?

Data:

• Length of wire (\(L\)) = \(12\text{ m}\)
• Area of cross-section (\(A\)) = \(4.0 \times 10^{-7}\text{ m}^2\)
• Resistance (\(R\)) = \(6.0\text{ }\Omega\)
• Resistivity (\(\rho\)) = ?

Solution:

According to the standard definition of electrical resistivity:

$$R = \rho \frac{L}{A} \implies \rho = \frac{R \cdot A}{L}$$

Plugging the parameters into our derived expression:

$$\rho = \frac{6.0 \times (4.0 \times 10^{-7})}{12}$$ $$\rho = \frac{2.4 \times 10^{-6}}{12} = 2.0 \times 10^{-7}\text{ }\Omega\cdot\text{m}$$

RESULT: The resistivity of the material at the temperature of the experiment is \(2.0 \times 10^{-7}\text{ }\Omega\cdot\text{m}\).

3. In a potentiometer arrangement, a cell of emf 1.20 V gives a balance point at 40.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 74.0 cm, what is the emf of the second cell?

Data:

• Emf of the primary cell (\(E_1\)) = \(1.20\text{ V}\)
• Initial balancing length (\(l_1\)) = \(40.0\text{ cm}\)
• Secondary balancing length (\(l_2\)) = \(74.0\text{ cm}\)
• Emf of the replacement cell (\(E_2\)) = ?

Solution:

In a balanced potentiometer configuration, the electromotive forces are directly proportional to their corresponding tracking balancing wire lengths:

$$\frac{E_2}{E_1} = \frac{l_2}{l_1} \implies E_2 = E_1 \times \frac{l_2}{l_1}$$

Calculating the final potential output:

$$E_2 = 1.20 \times \frac{74.0}{40.0}$$ $$E_2 = 1.20 \times 1.85 = 2.22\text{ V}$$

RESULT: The emf of the second cell is 2.22 V.

4. (a) Three resistors 1 \(\Omega\), 2 \(\Omega\), and 3 \(\Omega\) are combined in series. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 24 V and negligible internal resistance, obtain the potential drop across each resistor.

Data:

• Resistances: \(R_1 = 1\text{ }\Omega\), \(R_2 = 2\text{ }\Omega\), \(R_3 = 3\text{ }\Omega\)
• Battery source EMF (\(V\)) = \(24\text{ V}\)
• Equivalent Series Resistance (\(R_s\)) = ?
• Individual branch voltage drops (\(V_1, V_2, V_3\)) = ?

Solution:

Part (a): Since the components are linked end-to-end in series configuration:

$$R_s = R_1 + R_2 + R_3$$ $$R_s = 1 + 2 + 3 = 6\text{ }\Omega$$

Part (b): According to Ohm's Law, the total current output circulating through the main path is:

$$I = \frac{V}{R_s} = \frac{24}{6} = 4\text{ A}$$

Now, we evaluate the distinct individual drop values across every separate resistor element:

$$V_1 = I \cdot R_1 = 4\text{ A} \times 1\text{ }\Omega = 4\text{ V}$$ $$V_2 = I \cdot R_2 = 4\text{ A} \times 2\text{ }\Omega = 8\text{ V}$$ $$V_3 = I \cdot R_3 = 4\text{ A} \times 3\text{ }\Omega = 12\text{ V}$$

RESULT: The net total resistance is 6 \(\Omega\), and the individual potential drops across \(R_1\), \(R_2\), and \(R_3\) are 4V, 8V, and 12V respectively.

5. From the given circuit find the value of I.

Solution:

According to Kirchhoff’s Current Law (KCL), the algebraic sum of electrical current values entering a node point must strictly balance out the sum leaving that node point:

$$\sum I_{\text{incoming}} = \sum I_{\text{outgoing}}$$

By balancing the respective incoming node stream branches matching the problem system specification:

$$I = 0.6\text{ A}$$

RESULT: The total value of the targeted node current element I is 0.6 A.

6. In a meter bridge with a standard resistance of 15 \(\Omega\) in the right gap, the ratio of balancing length is 5:3. Find the value of the other resistance.

Data:

• Standard right gap resistance (\(S\)) = \(15\text{ }\Omega\)
• Balancing length division distribution ratio (\(l_1 : l_2\)) = \(5 : 3 \implies \frac{l_1}{l_2} = \frac{5}{3}\)
• Unknown left loop target resistance (\(R\)) = ?

Solution:

Using the analytical structural balanced condition for a traditional slide wire meter bridge system:

$$\frac{R}{S} = \frac{l_1}{l_2} \implies R = S \times \frac{l_1}{l_2}$$

Evaluating the mathematical values:

$$R = 15 \times \frac{5}{3} = 5 \times 5 = 25\text{ }\Omega$$

RESULT: The final computed alternative unknown value of resistance is 25 \(\Omega\).

7. By using KVL find current flowing through 10 ohm resistance.

Solution:

We systematically apply Kirchhoff’s Voltage Law (\(\sum V = 0\)) across the independent closed loops to build our linear system:

For Loop 1:

$$40I_1 - 20I_2 = 40 \implies 2I_1 - I_2 = 2 \quad \text{--- (i)}$$

For Loop 2:

$$-20I_1 + 60I_2 - 30I_3 = 0 \quad \text{--- (ii)}$$

For Loop 3:

$$-30I_2 + 40I_3 = -40 \implies -3I_2 + 4I_3 = -4 \quad \text{--- (iii)}$$

Let's map out our system parameters explicitly into determinants using Cramer's Rule:

$$\Delta = \begin{vmatrix} 40 & -20 & 0 \\ -20 & 60 & -30 \\ 0 & -30 & 40 \end{vmatrix} = 44000$$

Calculating the specialized substitution loop variable determinant for target current branch \(I_2\):

$$\Delta_2 = \begin{vmatrix} 40 & 40 & 0 \\ -20 & 0 & -30 \\ 0 & -40 & 40 \end{vmatrix} = 1400$$

Isolating the core loop variable values:

$$I_2 = \frac{\Delta_2}{\Delta} = \frac{1400}{44000} \approx 0.0318\text{ A}$$

RESULT: The exact current flowing through the shared 10 ohm resistance branch is 0.032 A (or 31.8 mA).