Question

A 5 V battery with internal resistance 1$\Omega$ and 2 V battery with internal resistance 2$\Omega$ are connected to a 5 $\Omega$ resistor as shown in the figure. The current in 5 $\Omega$ resistor is

Answer 1

12/17 A

Answer 2

The problem asks to find the current in the 5 $\Omega$ resistor in the given circuit. We can solve this using Kirchhoff's laws, specifically Kirchhoff's Current Law (KCL) and Ohm's law.

1. Define Nodes and Potentials:

Let's label the common top wire as node A and the common bottom wire as node B. We can assume the potential of node B to be 0 V (ground). Let the potential of node A be $V_A$.

2. Define Currents in Each Branch:

There are three parallel branches connected between node A and node B.

• Branch 1 (Left Branch): This branch contains a 5 V battery with an internal resistance of 1 $\Omega$. The positive terminal of the battery is connected towards node A, and the negative terminal towards node B. The current $I_1$ flowing from node A to node B through this branch can be expressed using Ohm's law. The effective voltage across this branch is $V_A - 5$ (if we consider current flowing from A to the battery's positive terminal, then through the battery and internal resistance to B). Or, more simply, the voltage drop across the resistor and battery is $V_A - 0 = V_A$. So, the current is $(V_A - 5)/1$ if we consider the battery as a voltage source in series with the resistor. Let's consider the current flowing out of the 5V battery towards node A. The potential drop across the 1$\Omega$ resistor is $I_1 \cdot 1$. So, $V_A = 5 - I_1 \cdot 1$. Thus, $I_1 = 5 - V_A$. This is the current flowing from the 5V battery towards node A.

• Branch 2 (Middle Branch): This branch contains only the 5 $\Omega$ resistor. The current $I_2$ flowing from node A to node B through this resistor is given by Ohm's law: $I_2 = \frac{V_A - 0}{5} = \frac{V_A}{5}$

• Branch 3 (Right Branch): This branch contains a 2 V battery with an internal resistance of 2 $\Omega$. Similar to Branch 1, let's consider the current flowing out of the 2V battery towards node A. The potential drop across the 2$\Omega$ resistor is $I_3 \cdot 2$. So, $V_A = 2 - I_3 \cdot 2$. Thus, $I_3 = \frac{2 - V_A}{2}$. This is the current flowing from the 2V battery towards node A.

3. Apply Kirchhoff's Current Law (KCL) at Node A:

According to KCL, the sum of currents entering a node is equal to the sum of currents leaving the node. Let's assume currents flowing into node A are positive and currents flowing out of node A are negative. Current from Branch 1 entering A: $I_1 = (5 - V_A)/1$ Current from Branch 3 entering A: $I_3 = (2 - V_A)/2$ Current from Branch 2 leaving A: $I_2 = V_A/5$

So, the KCL equation at node A is: $\frac{5 - V_A}{1} + \frac{2 - V_A}{2} = \frac{V_A}{5}$

4. Solve for $V_A$:

To eliminate the denominators, multiply the entire equation by the least common multiple of 1, 2, and 5, which is 10: $10 \left( \frac{5 - V_A}{1} \right) + 10 \left( \frac{2 - V_A}{2} \right) = 10 \left( \frac{V_A}{5} \right)$ $10(5 - V_A) + 5(2 - V_A) = 2V_A$ $50 - 10V_A + 10 - 5V_A = 2V_A$ $60 - 15V_A = 2V_A$ $60 = 2V_A + 15V_A$ $60 = 17V_A$ $V_A = \frac{60}{17} \text{ V}$

5. Calculate the Current in the 5 $\Omega$ Resistor:

The current in the 5 $\Omega$ resistor is $I_2$, which we defined as $\frac{V_A}{5}$. $I_2 = \frac{60/17}{5}$ $I_2 = \frac{60}{17 \times 5}$ $I_2 = \frac{12}{17} \text{ A}$

The decimal value is approximately $0.70588 \text{ A}$.

The final answer is $\boxed{\frac{12}{17} \text{ A}}$.