DC Circuits and Measurements — Practice Quiz

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.

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.

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.

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.