DC Circuit Analysis Questions and Answers

Answer: B

The work done is 20 J. The work done in moving a charge between two points is the charge multiplied by the potential difference.

\[ W = QV = 2 \times 10 = 20\ \text{J} \]

The value 5 J comes from dividing the voltage by the charge, 40 J from doubling the product, and 200 J from the product of 2, 10 and 10. So the second option is the answer. The joule is defined as one volt times one coulomb, so a potential difference of 1 V does 1 J of work on each coulomb moved across it.

Answer: C

The current is 40 mA. The power consumed in a resistor is P = I²R, so the current is found from the power and the resistance.

\[ I = \sqrt{\frac{P}{R}} = \sqrt{\frac{0.8}{500}} \]

\[ = \sqrt{0.0016} = 0.04\ \text{A} \]

That is 40 mA. The values 10 mA, 20 mA and 80 mA do not give 0.8 W in 500 Ω: for example 20 mA gives only 0.2 W, and 80 mA gives 3.2 W. So the third option is correct.

Answer: D

Germanium is not an insulator. It is a semiconductor, with a conductivity between that of a conductor and an insulator, and it is used for diodes and transistors.

Mica, glass and rubber are insulators and are used to prevent the flow of current. Germanium and silicon are the two common semiconductor materials, and germanium has a small energy gap of about 0.7 eV. The fourth option is therefore the one that is not an insulator.

Answer: C

Nichrome wire is an alloy of nickel and chromium (with some iron in some grades). It has a high resistivity and resists oxidation at high temperatures, so it is used for the heating elements of heaters, irons and toasters.

Nickel and aluminium, nickel and copper (a monel or constantan type alloy) and nickel and iron are other alloys with different uses. So the third option is correct.

Answer: D

The potential difference is 50 kV.

Potential difference is energy per unit charge, \(V = W/Q\).

\[ \begin{aligned} V &= \frac{50 \times 10^{-3}\ \text{J}}{1 \times 10^{-6}\ \text{C}} \\ &= 5 \times 10^{4}\ \text{V} = 50\ \text{kV} \end{aligned} \]

50 V comes from ignoring the difference between the milli- and micro- prefixes, and 5 V and 500 V misplace the power of ten in other ways.

Answer: A

One kWh is equal to about 860 kcal.

One kilowatt-hour is 1000 W for 3600 s, and one kilocalorie is about 4186 J.

\[ \begin{aligned} 1\ \text{kWh} &= 3.6 \times 10^{6}\ \text{J} \\ &= \frac{3.6 \times 10^{6}}{4186} \approx 860\ \text{kcal} \end{aligned} \]

This conversion is used constantly in heating problems, such as finding the energy needed to boil water. 765, 535 and 440 kcal do not correspond to any standard conversion.

Answer: D

A unit of specific resistance is the ohm-cm.

Specific resistance (resistivity) is defined by \(R = \rho l / A\), so \(\rho = RA/l\). Its units are ohm multiplied by area divided by length: ohm × cm² / cm = ohm-cm, or in SI the ohm-metre.

Ohm per cm is resistance per unit length, and ohm per square or cubic centimetre come from dividing by area or volume instead of multiplying by area and dividing by length.

Answer: B

The resistance will remain unchanged.

Resistance is \(R = \rho l / A\): it is proportional to length and inversely proportional to cross-sectional area. Doubling both gives

\[ R' = \rho \frac{2l}{2A} = R \]

Four times larger would need the length doubled and the area halved, and four times smaller the reverse. The change is fully determined by the formula, not random.

Answer: A

The power at 200 °C is about 611 W.

At constant supply voltage the power is \(V^{2}/R\), so it is inversely proportional to the resistance. With the temperature coefficient referred to 0 °C, \(R_{t} = R_{0}(1 + \alpha t)\).

\[ \begin{aligned} R_{800} &= R_{0}(1 + 0.32) = 1.32R_{0} \\ R_{200} &= R_{0}(1 + 0.08) = 1.08R_{0} \\ P_{200} &= 500 \times \frac{1.32}{1.08} = 611\ \text{W} \end{aligned} \]

The resistance is lower at the lower temperature, so the power must be higher than 500 W, which rules out 526 W as too small a change; 672 W and 620 W do not follow from the ratio of resistances.

Answer: D

The conductance is 0.5 mho.

Conductance is the reciprocal of resistance, \(G = 1/R\), measured in mho (siemens).

\[ G = \frac{1}{2\ \Omega} = 0.5\ \text{mho} \]

0.1, 0.3 and 0.4 mho would correspond to resistances of 10, 3.33 and 2.5 Ω.