Question

In the figure a part of circuit is shown:

• A. current will flow from A to B
• B. current may flow from A to B
• C. current will flow from B to A
• D. the direction of current will depend on r.

Answer 1

Answer: (B)

B

Answer 2

To determine the direction of current flow in the given circuit segment, we can apply Kirchhoff's Voltage Law (KVL) between points A and B.

Let $V_A$ be the potential at point A and $V_B$ be the potential at point B.

Given: $V_A = 20V$ $V_B = 2V$

There is a cell with EMF $\varepsilon$ and a resistor $r$ connected in series between A and B. The positive terminal of the cell is connected towards A, and the negative terminal is connected towards the resistor $r$ which then connects to B.

Let's assume a current $I$ flows from A to B. Applying KVL from point A to point B: Starting from $V_A$, we move across the cell. Since we are moving from the positive terminal to the negative terminal of the cell (in the assumed direction of current), there is a potential drop of $\varepsilon$. Then, we move across the resistor $r$ in the direction of the assumed current $I$. There is a potential drop of $I*r$.

So, the equation becomes: $V_A - \varepsilon - I*r = V_B$

Substitute the given potential values: $20V - \varepsilon - I*r = 2V$

Rearrange the equation to solve for $I$: $18 - \varepsilon = I*r$ $I = \frac{18 - \varepsilon}{r}$

Now, let's analyze the direction of current based on the value of $I$:

1. If $I > 0$: Current flows from A to B. This occurs when $\frac{18 - \varepsilon}{r} > 0$. Since $r$ (resistance) is always positive, this implies $18 - \varepsilon > 0$, or $\varepsilon < 18V$. So, if the EMF of the cell $\varepsilon$ is less than 18V, current will flow from A to B.

2. If $I < 0$: Current flows from B to A. This occurs when $\frac{18 - \varepsilon}{r} < 0$. Since $r$ is positive, this implies $18 - \varepsilon < 0$, or $\varepsilon > 18V$. So, if the EMF of the cell $\varepsilon$ is greater than 18V, current will flow from B to A.

3. If $I = 0$: No current flows. This occurs when $\frac{18 - \varepsilon}{r} = 0$. This implies $18 - \varepsilon = 0$, or $\varepsilon = 18V$. So, if the EMF of the cell $\varepsilon$ is exactly 18V, no current will flow.

Since the value of $\varepsilon$ is not provided in the problem, we cannot definitively state that current will flow in a particular direction (A to B or B to A). The direction depends on the value of $\varepsilon$.

Let's evaluate the given options: (A) current will flow from A to B: This is only true if $\varepsilon < 18V$. It's not always true. (B) current may flow from A to B: This is true, as it is possible if $\varepsilon < 18V$. (C) current will flow from B to A: This is only true if $\varepsilon > 18V$. It's not always true. (D) the direction of current will depend on r: This is incorrect. The direction of current depends on the sign of the net effective potential difference $(18 - \varepsilon)$, not on the value of $r$. The value of $r$ affects the magnitude of the current, but not its direction.

Therefore, the most appropriate option is (B), as it acknowledges a possibility that depends on the unknown value of $\varepsilon$.