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: Data Listing.
- Cell A: \( E_A = 2 \, \text{V} \), \( r_A = 5 \, \Omega \).
- Cell B: \( E_B = 3 \, \text{V} \), \( r_B = 5 \, \Omega \).
- Cell C: \( E_C = 5 \, \text{V} \), \( r_C = 1 \, \Omega \).
Cells are connected in parallel, meaning they share a common terminal voltage \( V \). The objective is to determine the currents \( I_A \), \( I_B \), and \( I_C \).Step 2: Current Equations.
The current for each cell is defined by the formula: \[I = \frac{E - V}{r}\]Applying this to each cell yields: \[I_A = \frac{2 - V}{5}, \quad I_B = \frac{3 - V}{5}, \quad I_C = \frac{5 - V}{1}\]As no external load is specified, Kirchhoff’s Current Law (KCL) at the junction states that the net current is zero: \[I_A + I_B + I_C = 0\]Step 3: Solve for Terminal Voltage \( V \).
Substituting the current equations into KCL: \[\frac{2 - V}{5} + \frac{3 - V}{5} + \frac{5 - V}{1} = 0\]Simplifying the equation: \[\frac{(2 - V) + (3 - V)}{5} + (5 - V) = 0\]\[\frac{5 - 2V}{5} + (5 - V) = 0\]Multiply by 5 to eliminate the fraction: \[(5 - 2V) + 5(5 - V) = 0\]\[5 - 2V + 25 - 5V = 0\]Combine terms: \[30 - 7V = 0\]Solving for \( V \): \[V = \frac{30}{7} \approx 4.2857 \, \text{V}\]Step 4: Current Calculation.
Using the calculated voltage \( V \), the currents are: - 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: Result Interpretation.
- \( I_A \approx -0.4571 \, \text{A} \): The negative sign indicates Cell A is being charged.
- \( I_B \approx -0.2571 \, \text{A} \): The negative sign indicates Cell B is being charged.
- \( I_C \approx 0.7143 \, \text{A} \): The positive sign indicates Cell C is supplying current.
Step 6: Verification.
Confirming KCL: \[-\frac{16}{35} - \frac{9}{35} + \frac{5}{7} = -\frac{25}{35} + \frac{25}{35} = 0\]The sum of currents is zero, validating the solution.% Final AnswerFinal Answer:
The calculated currents are: \[I_A = -\frac{16}{35} \, \text{A} \approx -0.4571 \, \text{A}, \quad I_B = -\frac{9}{35} \, \text{A} \approx -0.2571 \, \text{A}, \quad I_C = \frac{5}{7} \, \text{A} \approx 0.7143 \, \text{A}\]