Numerical Problems Based on EMF and Internal Resistance for Class 12 Physics

During charging, the current ($I$) flows into the positive terminal of the cell. The net driving emf is the difference between the source voltage and the cell’s emf.

Total resistance in the circuit $R_{total} = R + r = 5 + 1 = 6 \Omega$.

The terminal voltage ($V$) across the cell during charging is given by $V = E + Ir$:

The terminal potential difference ($V$) when current is drawn is given by $V = E – Ir$.

Case 1: $1.8 = E – 0.5r \quad \dots(1)$

Case 2: $1.6 = E – 1.0r \quad \dots(2)$

Subtracting equation (2) from equation (1):

Substitute the value of $r$ back into equation (1) to find $E$:

The potential difference of a cell in an open circuit (no current drawn) is equal to its electromotive force (emf).

When current $I = 2\text{ A}$ is drawn, the terminal voltage is $V = 4\text{ V}$. Using the equation $V = E – Ir$:

(i) EMF of the cell ($E$):
When the switch is open, no current flows. The ideal voltmeter measures the true emf of the cell.

(ii) Value of load resistance ($R$):
When the switch is closed, the voltmeter measures the terminal potential difference across $R$. Here, $V = 1.03\text{ V}$ and $I = 1.00\text{ A}$. By Ohm’s law ($V = IR$):

(iii) Internal resistance ($r$):
Using the equation for terminal voltage $V = E – Ir$:

Given $E = 6.0\text{ V}$, internal resistance $r = 1 \Omega$, and terminal voltage $V = 5.8\text{ V}$.

First, find the current ($I$) flowing through the circuit using the internal voltage drop ($E – V$):

Now, use Ohm’s law for the external resistor ($R$):

When a battery is being charged, the current enters the positive terminal, making the terminal potential difference greater than the emf.

The equation for charging is $V = E + Ir$, where $V = 7.2\text{ V}$, $E = 6.0\text{ V}$, and $I = 2.0\text{ A}$.

First, calculate the current ($I$) flowing through the circuit:

The total charge ($Q$) passing through in $t = 1\text{ s}$ is:

The number of electrons ($n$) can be found using the quantization of charge $Q = ne$, where $e = 1.6 \times 10^{-19}\text{ C}$:

Using the circuit equation $I = \frac{\varepsilon}{R + r}$ or $\varepsilon = I(R + r)$:

Case 1 ($I = 1\text{ A}, R = 4 \Omega$):

Case 2 ($I = 0.5\text{ A}, R = 9 \Omega$):

Equating (1) and (2):

Substituting $r$ back into equation (1):

True EMF $E = 4.0\text{ V}$.

When the voltmeter is connected, it draws a small current ($I$) because its resistance ($R_v$) is finite:

The voltmeter reading ($V$) is the potential difference across its terminals:

The error in the reading is $\Delta V = E – V$:

The percentage error is calculated relative to the true EMF:

Rounding to standard figures, the error is approximately $0.15\%$.

Given $E = 6\text{ V}$, internal resistance $r = 0.6 \Omega$, and external resistance $R = 2.4 \Omega$.

(a) Calculate current ($I$):

(b) Calculate potential difference ($V$):
The terminal voltage in the closed circuit can be found using either $V = E – Ir$ or $V = IR$. Using Ohm’s law for the external circuit: