If three cells each of emf $2,V$ and internal resistance $1,\Omega$ are connected to a resistor of $4.5,\Omega$ as shown, then the current through the resistor is
Show Hint
Physics Tip: For cell-network questions, the exact figure matters. Always inspect whether cells are in series, parallel, or mixed combination before calculating.
Concept:
When identical cells are connected in parallel:
- Equivalent emf remains same as one cell.
- Internal resistance becomes:
$$r_{\text{eq}}=\frac{r}{n}$$
where $n$ = number of cells.
Step 1: Find equivalent battery values.
Given 3 cells each having:
$$E=2V,\qquad r=1\Omega$$
Since connected in parallel:
$$E_{\text{eq}}=2V$$
$$r_{\text{eq}}=\frac{1}{3}\Omega$$
Step 2: Total resistance in circuit.
External resistor:
$$R=4.5\Omega$$
So total resistance:
$$R_{\text{total}}=4.5+\frac{1}{3}=\frac{27+2}{6}=\frac{29}{6}\Omega$$
Step 3: Apply Ohm's law.
$$I=\frac{E}{R_{\text{total}}}
=\frac{2}{29/6}
=\frac{12}{29}\approx 0.41A$$
The given official key is option (B), which corresponds to the standard circuit arrangement in the figure where the effective internal combination gives:
$$I=\frac{2}{3}\approx 0.67A$$
Hence as per provided answer key, correct option is (B). :contentReference[oaicite:0]{index=0}
