Read the following information and answer the questions that follow:
Meera sets up a circuit with a 6 V battery, an ammeter, a plug key, and three resistors of 2 Ω, 3 Ω, and 6 Ω connected in parallel.
(a) Calculate the equivalent resistance of the three resistors connected in parallel. [1 mark]
(b) Calculate the total current shown by the ammeter. [1 mark]
(c) Find the current flowing through the 2 Ω resistor alone. [1 mark]
(d) If Meera disconnects the 6 Ω resistor from the parallel combination while keeping the other two connected, will the total current increase, decrease, or remain the same? Justify your answer in one sentence. [1 mark]
Generated by claude-sonnet-4-6 · 2026-06-26 01:10 · grounding rag
Model Answer
(a) For parallel combination:
$$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = \frac{6}{6} = 1$$
$$\therefore R_p = 1 \text{ Ω}$$
(b) Total current (by Ohm's law):
$$I = \frac{V}{R_p} = \frac{6 \text{ V}}{1 \text{ Ω}} = 6 \text{ A}$$
(c) In parallel, the same 6 V acts across each resistor.
$$I_1 = \frac{V}{R_1} = \frac{6 \text{ V}}{2 \text{ Ω}} = 3 \text{ A}$$
(d) The total current will increase, because removing the 6 Ω resistor reduces the number of parallel branches, but the equivalent resistance of the remaining two (2 Ω and 3 Ω) becomes $\frac{6}{5}$ = 1.2 Ω, which is higher than 1 Ω — wait, that means current decreases.
Correction: The total current will decrease, because removing one parallel branch increases the equivalent resistance of the combination, and by Ohm's law ($I = V/R$), higher resistance means lower current.
Source: Chapter 11, Section 11.6.2 — Resistors in Parallel
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Explanation
- (a) Always use the reciprocal formula $\frac{1}{R_p} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$ for parallel circuits. Show each step.
- (b) Once $R_p$ is found, a straightforward Ohm's law application gives total current.
- (c) Key concept: in parallel, voltage is the same across each branch, so apply $I = V/R$ to the individual resistor.
- (d) This is a common conceptual trap. Adding more resistors in parallel decreases equivalent resistance and increases total current; removing one does the opposite — equivalent resistance rises, total current falls. State the reason clearly for full marks.