Two cells of emf 1 V and 2 V and internal resistance 2 and 1, res… — Solution | JEE Physics

NVAMediumJEE 2025Cells, EMF & Internal Resistance

Two cells of emf $1 \, \text{V}$ and $2 \, \text{V}$ and internal resistance $2 \, \Omega$ and $1 \, \Omega$ , respectively, are connected in series with an external resistance of $6 \, \Omega$ . The total current in the circuit is $I_1$ . Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is $I_2$ . The value of $\left( \frac{I_1}{I_2} \right)$ is $\frac{x}{3}$ . The value of x is _____ .

Answer

Correct answer:4

Step-by-step solution

Standard Method

Given: Two cells have emfs $1 \, \text{V}$ and $2 \, \text{V}$ with internal resistances $2 \, \Omega$ and $1 \, \Omega$ , and the external resistance is $6 \, \Omega$ .

Find: The value of $x$ if

\frac{I_1}{I_2} = \frac{x}{3}

For the series combination,

\varepsilon_{eq,s} = \varepsilon_1 + \varepsilon_2 = 1 + 2 = 3 \, \text{V}

r_{eq,s} = r_1 + r_2 = 2 + 1 = 3 \, \Omega

So the total resistance is

R_1 = r_{eq,s} + R = 3 + 6 = 9 \, \Omega

Hence,

I_1 = \frac{\varepsilon_{eq,s}}{R_1} = \frac{3}{9} = \frac{1}{3} \, \text{A}

For the parallel combination,

\varepsilon_{eq,p} = \frac{\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\frac{1}{2} + 2}{\frac{1}{2} + 1} = \frac{5/2}{3/2} = \frac{5}{3} \, \text{V}

r_{eq,p} = \frac{1}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{1}{\frac{1}{2} + 1} = \frac{2}{3} \, \Omega

So the total resistance is

R_2 = r_{eq,p} + R = \frac{2}{3} + 6 = \frac{20}{3} \, \Omega

Hence,

I_2 = \frac{\varepsilon_{eq,p}}{R_2} = \frac{5/3}{20/3} = \frac{1}{4} \, \text{A}

Now,

\frac{I_1}{I_2} = \frac{1/3}{1/4} = \frac{4}{3}

Given that

\frac{I_1}{I_2} = \frac{x}{3}

Comparing,

\frac{x}{3} = \frac{4}{3}

Therefore,

x = 4

The value of $x$ is $4$ .

Using equivalent cell formulas

Given: $E_1 = 1 \, \text{V}$ , $r_1 = 2 \, \Omega$ , $E_2 = 2 \, \text{V}$ , $r_2 = 1 \, \Omega$ , and $R = 6 \, \Omega$ .

Find: The value of $x$ from $\frac{I_1}{I_2} = \frac{x}{3}$ .

For cells in series,

E_{eq} = E_1 + E_2

r_{eq} = r_1 + r_2

Thus,

E_{eq,\,series} = 1 + 2 = 3 \, \text{V}

r_{eq,\,series} = 2 + 1 = 3 \, \Omega

Using complete circuit current,

I_1 = \frac{E_{eq,\,series}}{R + r_{eq,\,series}} = \frac{3}{6+3} = \frac{1}{3} \, \text{A}

For cells in parallel,

E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}

r_{eq} = \frac{r_1 r_2}{r_1 + r_2}

Hence,

E_{eq,\,parallel} = \frac{(1)(1) + (2)(2)}{2+1} = \frac{5}{3} \, \text{V}

r_{eq,\,parallel} = \frac{(2)(1)}{2+1} = \frac{2}{3} \, \Omega

Therefore,

I_2 = \frac{E_{eq,\,parallel}}{R + r_{eq,\,parallel}} = \frac{5/3}{6 + 2/3} = \frac{5/3}{20/3} = \frac{1}{4} \, \text{A}

Now evaluate the ratio:

\frac{I_1}{I_2} = \frac{1/3}{1/4} = \frac{4}{3}

Since

\frac{I_1}{I_2} = \frac{x}{3}

we get

x = 4

Therefore, the required numerical value is $4$ .

Common mistakes

Using the same equivalent emf formula for both configurations is incorrect. In series, emfs and internal resistances add directly, but in parallel the equivalent emf is a weighted combination. First identify the configuration, then apply the correct formula.
Ignoring internal resistance gives the wrong current values. The current must be calculated using the total circuit resistance, which is external resistance plus equivalent internal resistance. Always use $I = \frac{E_{eq}}{R + r_{eq}}$ for a complete circuit.
Taking the parallel internal resistance as $r_1 + r_2$ is wrong. For parallel cells, the equivalent internal resistance decreases and is found from reciprocal addition or $\frac{r_1 r_2}{r_1 + r_2}$ . Check whether the result is physically reasonable.

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