DC Circuits and Measurements — Important Questions
ICSE · Class 12 · Physics
42 important questions from DC Circuits and Measurements for ICSE Class 12 Physics, with answers. Includes multiple choice questions.
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Important Questions from DC Circuits and Measurements
n identical cells, each of EMF E and internal resistance r, are connected in parallel and supply current to an external resistance R. The current through R is maximum when:
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r >> R
Step 1: For n cells in parallel, I = nE/(r + nR). Step 2: If r >> R, then nR can be neglected: I ≈ nE/r. This is n times the current from a single cell — maximum benefit. Step 3: If r << R, then I ≈ nE/nR = E/R — same as a single cell, no benefit from parallel combination. Step 4: The parallel combination reduces equivalent internal resistance to r/n, which is most beneficial when this reduction matters — i.e., when r is large compared to R. Step 5: Parallel combination is useful for high internal resistance cells with small external load.
A metre bridge shows null deflection at 40 cm when an unknown resistance X is in one gap and a 6 Ω resistance in the other. If X and the 6 Ω are interchanged, the new null point is at:
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60 cm
Step 1: Original condition: X/6 = l/(100−l) = 40/60 = 2/3, so X = 4 Ω. Step 2: After interchanging, the 6 Ω is in the left gap and X (= 4 Ω) is in the right gap. Step 3: New balance condition: 6/4 = l′/(100−l′). Step 4: 6(100−l′) = 4l′ → 600 = 10l′ → l′ = 60 cm. Step 5: Note: when X and known resistance are interchanged, the new null point is at (100 − old null point) ONLY when X = known resistance. Here they differ, so we must calculate properly.
A cell is being charged by an external source. If the EMF of the cell is 6 V, internal resistance is 1 Ω, and the charging current is 2 A, what is the terminal voltage of the cell during charging?
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8 V
Step 1: During charging, current is forced into the cell against its EMF (from + to − inside the cell). Step 2: The terminal voltage during charging is V = E + Ir (not E − Ir). Step 3: V = 6 + (2)(1) = 8 V. Step 4: This is higher than the EMF because the external source must overcome both the cell's EMF and its internal resistance. Step 5: During discharging V = E − Ir (terminal voltage is less than EMF). This distinction is critical — the sign of Ir reverses depending on whether cell is charging or discharging.
In a circuit, two cells of EMF 4 V (internal resistance 1 Ω) and 2 V (internal resistance 1 Ω) are connected in series opposing each other (positive terminals facing each other) and drive current through a 4 Ω external resistance. What is the current in the circuit?
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0.33 A
Step 1: When two cells oppose each other in series, net EMF = E₁ − E₂ = 4 − 2 = 2 V. Step 2: Total internal resistance = r₁ + r₂ = 1 + 1 = 2 Ω (internal resistances always add in series regardless of cell orientation). Step 3: Total resistance = 2 + 4 = 6 Ω. Step 4: Current I = Net EMF / Total resistance = 2/6 = 1/3 ≈ 0.33 A. Step 5: Current flows in the direction of the stronger cell (4 V cell). The 2 V cell is being charged in this arrangement.
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