In the circuit shown in figure E1 = 3 V, E2 = 2 V, E3 = 1 V and R = r1 = r2 = r3 = 1 ohm.
(a) Find t

Q: Why is the current through resistor R zero in part (a)?

In part (a), the terminals A and B are open, meaning there is no closed path for the current to flow through R and return to the circuit. Without a complete loop, the current must be zero.

Q: How do we determine the direction of the currents?

We can assume any direction initially. Here, we assumed the currents flow from right to left (from B to C) because the positive terminals of the batteries are on the left. If a calculated current is negative, it simply means the actual flow is in the opposite direction.

Q: Why does short-circuiting $r_2$ fix the voltage across all branches?

When $r_2$ is zero, the middle branch becomes an ideal voltage source with no internal voltage drop. In a parallel circuit, all branches share the same potential difference. Therefore, the ideal battery $E_2$ forces the entire parallel combination to operate at its exact EMF.

Q: What happens to the equivalent EMF formula in part (b)?

The standard equivalent EMF formula involves dividing by the internal resistances. If one resistance is zero, that term becomes infinite, which mathematically shows that the branch with zero resistance completely dominates and sets the voltage for the entire parallel group.

Q: Why do we use Kirchhoff's Current Law (KCL) at junction C in part (b)?

While we can easily find the currents in branches 1, 3, and R using Ohm's law and the clamped voltage, branch 2 has zero resistance, so we cannot use $I = V/R$ for it. Instead, we use KCL to find $i_2$ by ensuring the total current entering junction C equals the total current leaving it.

Question

In the circuit shown in figure $E_1 = 3	ext{ V}$, $E_2 = 2	ext{ V}$, $E_3 = 1	ext{ V}$ and $R = r_1 = r_2 = r_3 = 1\ \Omega$.
(a) Find the potential difference between the points $A$ and $B$ and the currents through each branch.
(b) If $r_2$ is short-circuited and the point $A$ is connected to point $B$, find the currents through $E_1, E_2, E_3$ and the resistor $R$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Let's begin by carefully examining the circuit for part (a). The terminals $A$ and $B$ are completely open. Because there is no closed loop connecting these two points externally, absolutely no current can flow through the resistor $R$.

This is a crucial observation. Since the current through $R$ is zero, there is no voltage drop across it ($V = IR = 0$). This implies that the electrical potential at point $A$ is exactly the same as the potential at the junction $C$. Therefore, finding the potential difference $V_{AB}$ is equivalent to finding the potential difference $V_{CB}$ across the parallel branches.

### The Master Equation

Between junction $C$ and point $B$, we have three distinct branches connected in parallel. Each branch contains a battery and an internal resistance. To simplify this complex network, we can replace the entire parallel combination with a single equivalent battery using the standard formula for parallel EMFs:

$$E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2} + \frac{E_3}{r_3}}{\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}}$$

Substituting the given values ($E_1=3	ext{ V}, E_2=2	ext{ V}, E_3=1	ext{ V}$ and $r_1=r_2=r_3=1\ \Omega$), we get:

$$E_{eq} = \frac{\frac{3}{1} + \frac{2}{1} + \frac{1}{1}}{\frac{1}{1} + \frac{1}{1} + \frac{1}{1}} = \frac{6}{3} = 2	ext{ V}$$

### Calculating the Initial Currents

Since the external circuit is open, the potential difference across the parallel combination is simply the equivalent EMF. Thus, $V_{CB} = 2	ext{ V}$, which means $V_{AB} = 2	ext{ V}$.

Now, we can find the current in each individual branch. We will assume the current flows from right to left (from $B$ to $C$) because the positive terminals of the batteries are on the left. The voltage equation for any branch is $V_{CB} = E - ir$.

For branch 1: 
$$2 = 3 - i_1(1) \Rightarrow i_1 = 1	ext{ A}$$

For branch 2: 
$$2 = 2 - i_2(1) \Rightarrow i_2 = 0	ext{ A}$$

For branch 3: 
$$2 = 1 - i_3(1) \Rightarrow i_3 = -1	ext{ A}$$

The negative sign for $i_3$ simply indicates that the actual current flows in the opposite direction (from left to right).

### Modifying the Circuit

In part (b), the circuit undergoes a dramatic transformation. The resistor $r_2$ is short-circuited, meaning its resistance drops to exactly $0\ \Omega$. Furthermore, point $A$ is connected directly to point $B$.

By connecting $A$ to $B$, the resistor $R$ is now placed directly in parallel with the three battery branches across points $C$ and $B$.

### The Power of an Ideal Source

The short circuit in the middle branch is the key to solving this part effortlessly. Because $r_2 = 0$, the middle branch acts as an **ideal voltage source**. An ideal voltage source connected in parallel with other components will forcefully clamp the potential difference across the entire parallel combination to its own EMF.

Therefore, the potential difference between $C$ and $B$ is rigidly fixed at $E_2$:

$$V_{CB} = 2	ext{ V}$$

### Final Calculation

With the voltage firmly established at $2	ext{ V}$, we can easily calculate the new currents. The current through the external resistor $R$ is:

$$i_R = \frac{V_{CB}}{R} = \frac{2}{1} = 2	ext{ A}$$

For branch 1, the current remains unchanged because the voltage across it is still $2	ext{ V}$:

$$2 = 3 - i_1(1) \Rightarrow i_1 = 1	ext{ A}$$

Similarly, for branch 3:

$$2 = 1 - i_3(1) \Rightarrow i_3 = -1	ext{ A}$$

Finally, to find the current $i_2$ flowing through the ideal battery, we apply Kirchhoff's Current Law (KCL) at junction $C$. The sum of the currents entering the junction from the batteries must equal the current leaving through resistor $R$:

$$i_1 + i_2 + i_3 = i_R$$

$$1 + i_2 - 1 = 2 \Rightarrow i_2 = 2	ext{ A}$$

And there we have it! A seemingly complex circuit elegantly unraveled by understanding the behavior of parallel EMFs and ideal voltage sources.

Visualized Solution

$V_{A} = V_{C}$

$E_{e q} = \frac{\sum \frac{E _{i}}{r _{i}}}{\sum \frac{1}{r _{i}}}$

$E_{e q} = 2 V$

$V_{A B} = 2 V$

$i_{1}, i_{2}, i_{3}$

$r_{2} = 0$

$V_{C} - V_{B} = 2 V$

$i_{1} + i_{2} + i_{3} = i_{R}$

The Sigma Insight: Cells, EMF, and Internal Resistance

Analyzing the Setup

The Master Equation

Calculating the Initial Currents

Modifying the Circuit

The Power of an Ideal Source

Final Calculation

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Visualized Solution

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