Here R2,R3,R4 are in parallel $\begin{aligned}
& \frac{1}{\mathrm{R}{234}}=\frac{1}{\mathrm{R}2}+\frac{1}{\mathrm{R}3}+\frac{1}{\mathrm{R}4} \
& \mathrm{R}{234}=2 \Omega
\end{aligned}R{234}isinserieswithR_1so\begin{aligned}
& \mathrm{R}{\mathrm{eq}}=\mathrm{R}{234}+\mathrm{R}_1=2+10=12 \Omega \
& \mathrm{i}=\frac{12}{12}=1 \mathrm{Amp}
\end{aligned}$
JEE Main 2024 — Physics Electromagnetism
2024mcqmedium
In the given figure R1=10Ω,R2=8Ω,R3=4Ω and R4=8Ω. Battery is ideal with emf 12 V. Equivalent resistant of the circuit and current supplied by battery are respectively : 
Official previous-year question
Held on 5 Apr 2024 · Verified 6 Jul 2026.
Options
- A
10.5Ω and 1.14 A
- B
12Ω and 1 A
- C
10.5Ω and 1 A
- D
12Ω and 11.4 A
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