Two batteries of emf 4 V and 8 V with internal resistance 1 Ω and 2 Ω are connected in a circuit with resistance of 9 Ω as shown in figure. The current and potential difference between the points P and Q are (a) 1/3 A and 3 V (b) 1/6 A and 4 V (c) 1/9 A and 9 V (d) 1/12 A and 12 V

Answer: (a). The two cells have their + terminals facing each other, so they oppose: net emf = 8 - 4 = 4 V. Total resistance = 1 + 2 + 9 = 12 Ω I = 4/12 = 1/3 A VPQ = drop across R = 1/3 × 9 = 3 V So the answer is (a) 1/3 A and 3 V.

Two batteries of emf 4 V and 8 V with internal resistance 1 Ω and 2 Ω… | NEET 1988 (AIPMT)

Q210NEET 1988 · AIPMTMedium

Kirchhoff's Laws & Measuring Instruments

Two batteries of emf $4\text{ V}$ and $8\text{ V}$ with internal resistance $1\ \Omega$ and $2\ \Omega$ are connected in a circuit with resistance of $9\ \Omega$ as shown in figure. The current and potential difference between the points $P$ and $Q$ are

Show answer and solution

Answer: (a)

The two cells have their + terminals facing each other, so they oppose: net emf $= 8 - 4 = 4\text{ V}$ .
Total resistance $= 1 + 2 + 9 = 12\ \Omega$
$I = \frac{4}{12} = \frac{1}{3}\text{ A}$
$V_{PQ}$ = drop across $R$ $= \frac{1}{3} \times 9 = 3\text{ V}$
So the answer is (a) $\frac{1}{3}$ A and 3 V.

Report a mistake

Kirchhoff's Laws & Measuring Instruments question from NEET 1988 (AIPMT)

More questions on Kirchhoff's Laws & Measuring Instruments