DC Circuit Analysis Questions and Answers

Answer: A

An ideal voltage source has zero internal resistance.

An ideal voltage source keeps the same terminal voltage whatever current the load draws. Any internal resistance would cause a voltage drop that grows with the current, so the ideal source must have none.

Infinite internal resistance is the property of an ideal current source. The size of the emf and the amount of current drawn do not make a source ideal.

Answer: C

The new resistance is 8R Ohms.

Resistance is \(R = \frac{\rho l}{A}\) with \(A = \frac{\pi d^{2}}{4}\), so \(R \propto \frac{l}{d^{2}}\).

\[ \begin{aligned} \frac{R'}{R} &= \frac{2l}{l} \times \left(\frac{d}{d/2}\right)^{2} \\ &= 2 \times 4 = 8 \end{aligned} \]

Halving the diameter quarters the area, which alone makes the resistance four times larger, and doubling the length doubles it again. 4R forgets the length, and 2R treats the area as proportional to the diameter instead of its square.

Answer: B

The equivalent resistance is 2 Ω.

For n equal resistances R in parallel, each branch adds the same conductance 1/R, so the total conductance is n/R and the equivalent resistance is R/n.

\[ R_{eq} = \frac{R}{n} = \frac{32}{16} = 2\ \Omega \]

A parallel combination is always smaller than its smallest branch, which rules out 32 Ω, 16 Ω and 1024 Ω. 32 Ω is a single branch, and the series sum of sixteen 32 Ω resistors would be 512 Ω.

Answer: B

Only the 100 W lamp will fuse.

Find each lamp's resistance from its rating, \(R = V^{2}/P\), then the series current and the share of the 440 V across each lamp.

\[ \begin{aligned} R_{100} &= \frac{220^{2}}{100} = 484\ \Omega \\ R_{200} &= \frac{220^{2}}{200} = 242\ \Omega \\ I &= \frac{440}{484 + 242} = 0.606\ \text{A} \end{aligned} \]

\[ \begin{aligned} V_{100} &= 0.606 \times 484 = 293\ \text{V} \\ V_{200} &= 0.606 \times 242 = 147\ \text{V} \end{aligned} \]

The 100 W lamp, having the higher resistance, takes 293 V, well above its 220 V rating, and burns out. Once it fails the series circuit opens and the 200 W lamp, which was only getting 147 V, is left intact, so neither 'both' nor 'only the 200 W lamp' is right.

Answer: A

Appliances are connected in parallel because it makes the operation of appliances independent of each other.

In a parallel circuit every appliance is connected directly across the supply and gets the full supply voltage it is rated for. Each one can be switched on or off without affecting the others, and if one fails the rest keep working. In a series circuit one switch or one failure would cut off everything, and the voltage would be shared out between the appliances.

A parallel circuit actually draws more total current, since the branch currents add, and it does not reduce power loss. It is the independence of the appliances, not simplicity, that decides the choice.

Answer: B

One coulomb is the charge on \(625 \times 10^{16}\) electrons.

Each electron carries \(1.6 \times 10^{-19}\) C, so the number in one coulomb is the reciprocal.

\[ \begin{aligned} n &= \frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18} \\ &= 625 \times 10^{16} \end{aligned} \]

The option is simply \(6.25 \times 10^{18}\) written with a different power of ten. \(6.25 \times 10^{16}\) drops a factor of 100, and the other two options are out by factors of 10 and 1000.