Numerical Problems Based on Combination of Cells for Class 12 Physics

For $n = 3$ identical cells in series:

Total EMF ($E_{eq}$) $= nE = 3 \times 2 = 6\text{ V}$.

Total internal resistance ($r_{eq}$) $= nr = 3 \times 0.2 = 0.6 \Omega$.

The total resistance of the circuit is the sum of external resistance ($R$) and total internal resistance:

The current ($I$) in the circuit is:

For $m = 3$ identical cells connected in parallel, the equivalent EMF is the same as a single cell: $E_{eq} = 2\text{ V}$.

The equivalent internal resistance is $r_{eq} = \frac{r}{3}$.

Given the terminal voltage $V = 1.5\text{ V}$ across the external resistor $R = 5 \Omega$, the total current ($I$) is:

Using the terminal voltage equation $V = E_{eq} – I r_{eq}$:

Total EMF for series connection: $E_{eq} = E_1 + E_2 = 1.5 + 1.5 = 3.0\text{ V}$.

Total internal resistance: $r_{eq} = r_1 + r_2 = 0.5 + 0.25 = 0.75 \Omega$.

1. Calculate current ($I$):

2. Calculate potential difference across each cell:

For the first cell ($V_1$):

For the second cell ($V_2$):

Let the EMF of each cell be $E$ and internal resistance be $r$. Number of cells $n = 10$.

Case 1 (Series): Equivalent EMF $= 10E$, equivalent internal resistance $= 10r$.

Case 2 (Parallel): Equivalent EMF $= E$, equivalent internal resistance $= \frac{r}{10}$.

Substitute equation (2) into equation (1):

Now, calculate $E$ using equation (2):

For a mixed grouping of cells with $n$ cells in series per row and $m$ rows in parallel, the current is:

Given $I = 1.5\text{ A}$, $E = 1.5\text{ V}$, $R = 30 \Omega$, and $r = 1.0 \Omega$:

For minimum cells (minimum $N = nm$), we apply the maximum power condition where the external resistance equals the equivalent internal resistance:

Substitute $n = 30m$ into the current equation:

Since $m = 2$, we find $n$:

Total minimum cells required $N = n \times m = 60 \times 2 = 120\text{ cells}$.

Let the EMF of each cell be $E$ and internal resistance be $r$. External resistance $R = 1 \Omega$.

Current in series combination ($I_s$):

Current in parallel combination ($I_p$):

Given that $I_s = I_p$:

Cross-multiplying to solve for $r$:

Since $R = 1 \Omega$, the internal resistance $r = 1 \Omega$.

This is a mixed grouping with $n = 2$ cells per row and $m = 2$ rows. $E = 2\text{ V}$, $r = 1.5 \Omega$, and $R = 10 \Omega$.

Equivalent EMF of the setup ($E_{eq}$):

Equivalent internal resistance ($r_{eq}$):

1. Calculate the total current and branch current:

By rounding to standard significant decimal places used in textbook answers, $I_{total} \approx 0.35\text{ A}$. The current divides equally among the $m = 2$ parallel branches:

2. Calculate the potential difference ($V$) across the $10 \Omega$ load: