Question

Three cells A, B, and C of EMFs 2 V, 3 V, and 5 V respectively are connected in parallel to each other. Their internal resistances are 5 \( \Omega \), 5 \( \Omega \), and 1 \( \Omega \) respectively. Calculate the currents flowing through the cells A, B, and C.

Hint:

In parallel cells with different EMFs, the cell with the highest EMF supplies current, while others may be charged if their EMF is less than the terminal voltage.

Answer 1

Step 1: List the given data.
- Cell A: \( E_A = 2 \, \text{V} \), \( r_A = 5 \, \Omega \).
- Cell B: \( E_B = 3 \, \text{V} \), \( r_B = 5 \, \Omega \).
- Cell C: \( E_C = 5 \, \text{V} \), \( r_C = 1 \, \Omega \).
The cells are in parallel, so they share the same terminal voltage \( V \). We need to find the currents \( I_A \), \( I_B \), and \( I_C \). Step 2: Set up the current equations.
The current through each cell is: \[ I = \frac{E - V}{r} \] \[ I_A = \frac{2 - V}{5}, \quad I_B = \frac{3 - V}{5}, \quad I_C = \frac{5 - V}{1} \] Since no external load is mentioned, apply Kirchhoff’s Current Law (KCL) at the junction (net current is zero): \[ I_A + I_B + I_C = 0 \] Step 3: Solve for \( V \).
\[ \frac{2 - V}{5} + \frac{3 - V}{5} + \frac{5 - V}{1} = 0 \] \[ \frac{(2 - V) + (3 - V)}{5} + (5 - V) = 0 \] \[ \frac{5 - 2V}{5} + (5 - V) = 0 \] \[ (5 - 2V) + 5(5 - V) = 0 \] \[ 5 - 2V + 25 - 5V = 0 \] \[ 30 - 7V = 0 \] \[ V = \frac{30}{7} \approx 4.2857 \, \text{V} \] Step 4: Calculate the currents.
- Cell A: \[ I_A = \frac{2 - \frac{30}{7}}{5} = \frac{\frac{14 - 30}{7}}{5} = \frac{-\frac{16}{7}}{5} = -\frac{16}{35} \approx -0.4571 \, \text{A} \] - Cell B: \[ I_B = \frac{3 - \frac{30}{7}}{5} = \frac{\frac{21 - 30}{7}}{5} = \frac{-\frac{9}{7}}{5} = -\frac{9}{35} \approx -0.2571 \, \text{A} \] - Cell C: \[ I_C = \frac{5 - \frac{30}{7}}{1} = \frac{\frac{35 - 30}{7}}{1} = \frac{5}{7} \approx 0.7143 \, \text{A} \] Step 5: Interpret the results.
- \( I_A \approx -0.4571 \, \text{A} \): Negative, so Cell A is being charged.
- \( I_B \approx -0.2571 \, \text{A} \): Negative, so Cell B is being charged.
- \( I_C \approx 0.7143 \, \text{A} \): Positive, so Cell C supplies current. Step 6: Verify.
\[ -\frac{16}{35} - \frac{9}{35} + \frac{5}{7} = -\frac{25}{35} + \frac{25}{35} = 0 \] The sum is zero, confirming the solution. % Final Answer Final Answer:
The currents are: \[ I_A = -\frac{16}{35} \, \text{A} \approx -0.4571 \, \text{A}, \quad I_B = -\frac{9}{35} \, \text{A} \approx -0.2571 \, \text{A}, \quad I_C = \frac{5}{7} \, \text{A} \approx 0.7143 \, \text{A} \]