Understanding Series Circuits and Ohm's Law

In this problem, we have a simple electrical circuit consisting of a cell providing an electromotive force (e.m.f.) and three resistors connected in series. We need to determine the potential difference across one of the resistors, specifically the 3 ohm resistor. To solve this, we will use the fundamental principles of series circuits and Ohm's Law.

Calculating Total Resistance in a Series Circuit

When resistors are connected in series, the total resistance of the circuit is the sum of the individual resistances. The resistors given are 2 ohm, 3 ohm, and 5 ohm.

Let $R_1 = 2 \, \Omega$, $R_2 = 3 \, \Omega$, and $R_3 = 5 \, \Omega$.

The total resistance ($R_{total}$) is:

$\qquad R_{total} = R_1 + R_2 + R_3$

$\qquad R_{total} = 2 \, \Omega + 3 \, \Omega + 5 \, \Omega$

$\qquad R_{total} = 10 \, \Omega$

So, the total resistance of the series combination is 10 ohm.

Calculating Total Current in the Circuit

The cell provides an e.m.f. of 2 volts, and its internal resistance is negligible. This means the total voltage across the external circuit is equal to the e.m.f. of the cell, which is 2 volts.

According to Ohm's Law, the current ($I$) flowing through a circuit is given by the voltage ($V$) divided by the total resistance ($R$).

$\qquad I = \frac{V}{R_{total}}$

Here, $V = 2 \, V$ and $R_{total} = 10 \, \Omega$.

$\qquad I = \frac{2 \, V}{10 \, \Omega}$

$\qquad I = 0.2 \, A$

In a series circuit, the same current flows through every component, including each resistor. Therefore, a current of 0.2 A flows through the 3 ohm resistor.

Calculating Potential Difference Across the 3 ohm Resistor

To find the potential difference ($V_{3\Omega}$) across the 3 ohm resistor ($R_2$), we use Ohm's Law again, applied specifically to this resistor:

$\qquad V_{3\Omega} = I \times R_2$

We know $I = 0.2 \, A$ and $R_2 = 3 \, \Omega$.

$\qquad V_{3\Omega} = 0.2 \, A \times 3 \, \Omega$

$\qquad V_{3\Omega} = 0.6 \, V$

The potential difference across the 3 ohm resistance is 0.6 volts.

Summary of Calculations

Conclusion

Based on the calculations using series circuit rules and Ohm's Law, the potential difference across the 3 ohm resistor is 0.6 V.

Revision Table: Key Concepts

Additional Information: Potential Difference in Series Circuits

In a series circuit, the potential difference across each resistor is generally different, depending on the value of the resistance. The total potential difference across the combination of resistors is equal to the sum of the potential differences across each individual resistor. This is a consequence of Kirchhoff's Voltage Law.

For this circuit, we could also calculate the potential difference across the other resistors:

• Potential difference across 2 $\Omega$: $V_{2\Omega} = I \times R_1 = 0.2 \, A \times 2 \, \Omega = 0.4 \, V$
• Potential difference across 5 $\Omega$: $V_{5\Omega} = I \times R_3 = 0.2 \, A \times 5 \, \Omega = 1.0 \, V$

The sum of these potential differences is $0.4 \, V + 0.6 \, V + 1.0 \, V = 2.0 \, V$. This sum equals the total voltage supplied by the cell (2V), confirming our calculations are consistent with the principles of series circuits and Kirchhoff's Voltage Law.