DC Circuit Analysis Questions and Answers

Answer: C

The branch ac, which carries the galvanometer, has no current when the bridge is balanced. The balance condition is 60 / 60 = 40 / R, so R is 40 ohms. Each side of the bridge is then 60 + 40 = 100 ohms (b to a to d, and b to c to d), and the two sides are in parallel, so the bridge resistance is 50 ohms.

\[ I_{B} = \frac{2}{50} = 0.04\ \mathrm{A} = 40\ \mathrm{mA} \]

So the battery current is 40 mA, option (C). The values 2 mA, 10 mA and 20 mA would need a larger resistance.

Answer: D

Trickle charging supplies a continuous constant current at a low rate to a battery that is already charged, only enough to make up for its self-discharge, so it is kept in a fully charged condition. Freshening charging is given to a battery that has stood idle, boost charging is a quick charge at a high rate, and initial charging is the first charge given to a new battery. The answer is trickle charging, option (D).

Answer: B

Each cell is 1.5 V and 2 AH. To get 6 V the cells must be in series, and 6 / 1.5 = 4 cells per row. To get 4 AH the rows must be in parallel, since the capacity adds, and 4 / 2 = 2 rows. So 4 series cells and 2 rows are needed, which is 8 cells in all, option (B). The other options give the wrong voltage or capacity: 2 series cells would give only 3 V, 8 series cells would give 12 V, and 8 rows would give 16 AH.

Answer: A

The resistors R1, R2 and R3 are in series across the 200 V supply, so the voltage drops add up to the supply voltage.

\[ V_{3} = 200 - 60 - 60 = 80\ \mathrm{V} \]

So the voltage across R3 is 80 V, option (A). The values 40 V, 20 V and 120 V would not add up with 60 V and 60 V to the 200 V of the supply. The same rule, that the voltage drops in a series circuit add up to the supply voltage, is Kirchhoff's voltage law. Here R1 and R2 each take 60 V, so together they take 120 V, and the 80 V left over must fall across R3. Since the current is the same in all three resistors, R3 has the largest resistance of the three, because its drop is the largest.

Answer: B

In the colour code of a resistor the first band gives the first digit and the second band the second digit, the third band is the multiplier, and the fourth band is the tolerance. Brown is 1, violet is 7 and orange is a multiplier of 1000, so the value is 17 x 1000 = 17,000 ohms. A gold fourth band means a tolerance of plus or minus 5 percent. So the resistor is 17,000 ohms with plus or minus 5 percent tolerance, option (B). The other values do not match the bands brown, violet and orange.

Answer: D

The three capacitors are connected in parallel between the terminals a and b, and in a parallel connection the capacitances add up, since the plate areas add.

\[ C_{ab} = 15 + 25 + 25 = 65\ \mathrm{F} \]

So the equivalent capacitance is 65 F, option (D). The values 15 F, 27.5 F and 35 F would result from combining the capacitors in series or from adding only some of them; in series the reciprocals would add instead. In a series connection the equivalent capacitance would be smaller than the smallest of the three, 15 F, so the answers of 15 F and below belong to a series arrangement.

Answer: D

Capacitors connected in parallel all have the same voltage across them, so the voltage applied to the group cannot be more than the lowest breakdown voltage of any one of them. Two of the capacitors can take 250 V but one can take only 200 V, so the highest voltage that can be applied without damaging any of them is 200 V, option (D). The values 250 V, 600 V and 650 V would break down the 200 V capacitor, and the voltages do not add in parallel.

Answer: D

The mercury cell is a small button cell with a very steady voltage and a long life for its size, so it was the cell most commonly used in miniature equipment such as digital watches, hearing aids and calculators, as the key gives. A voltaic cell is a simple cell of little practical use, and the Leclanche cell is a bulky dry or wet cell used in torches and bells, and the alkaline cell is larger and in the key's view not the one for watches. The answer is Mercury cell, option (D).

Answer: D

The effective resistance is 150 Ω. In a series connection the resistances add.

\[ R = 50 + 100 = 150\ \Omega \]

The values 25 Ω, 50 Ω and 100 Ω are not the sum of 50 Ω and 100 Ω, and 100 Ω is simply one of the two resistors. None of these is the total resistance, so the fourth option is the answer. In a parallel connection the total would be less than the smallest resistor, which is not the case here.

Answer: C

The effective resistance is 100 Ω. For two equal resistors in parallel, the total is half of one resistor.

\[ R = \frac{200 \times 200}{200 + 200} = 100\ \Omega \]

The values 25 Ω and 50 Ω are too low for two 200 Ω resistors in parallel, and 150 Ω is not what a parallel combination gives (the result of a parallel connection is always less than the smaller resistor, 200 Ω). So the third option is correct.