Ohm’s Law, KCL, KVL and Dependent Sources: Practice Problems with Answers

S1Which connections are allowed?Two-tier

Each panel joins two ideal sources. Which connection is not permitted?

Answer

a)(a)b)(b)c)(c)d)(d)

Reason

i)Two ideal current sources in one loop must carry the same current, so 2 A and 5 A conflict.ii)An ideal voltage source can never be connected in parallel with an ideal current source.iii)Two ideal voltage sources can never be connected in parallel.iv)The 5 A source would overload the 2 A source.

S2Reading a dependent source

A source drawn as a diamond containing an arrow is labelled $0.4v_x$, where $v_x$ is the voltage across a resistor elsewhere in the circuit. Which description is correct?

Answer

a)A current-controlled voltage source; 0.4 is in V/Ab)A voltage-controlled current source; 0.4 is in A/Vc)A voltage-controlled voltage source; 0.4 has no unitd)An independent current source of 0.4 A

S3Ohm’s law and the reference arrowTwo-tier

Find the current $i$.

Answer

a)3 Ab)−3 Ac)75 Ad)−75 A

Reason

i)The arrow enters the + terminal, so $v = iR$.ii)The arrow enters the − terminal, so $v = -iR$.iii)Current flows from + to −, so $i$ is positive.iv)$i = v/R$ for every resistor, whatever the arrow.

Follow-up: through the resistor, does positive charge actually move from a to b or from b to a?

a)from a to bb)from b to a

S4Power in that resistorTwo-tier

For the resistor in S3, the power is:

Answer

a)45 W absorbedb)45 W deliveredc)75 W absorbedd)0 W

Reason

i)$p = vi = (15)(-3) = -45$ W, so it delivers.ii)The arrow enters the − terminal, so $p = -vi = 45$ W; a resistor always absorbs, $p = i^2R$.iii)$i$ is negative, so no power flows.iv)$p = v^2R$.

S5Resistor ratings

A 1 kΩ resistor is rated ¼ W. What is the largest voltage you can put across it without exceeding its rating?

Answer

a)250 Vb)15.8 Vc)0.25 Vd)4 V

S6A heater on your generator linePredict first

An electric heater is rated 2000 W at 220 V.

a) Find its resistance

Ω

a) … and the current it draws

A

b) Can it run on a 5 A generator subscription?

a)Yesb)No

c) Predict first: if the line voltage sags to 200 V (9% lower), does the heater’s power fall by less than 9%, about 9% or more than 9%?

a)Less than 9%b)About 9%c)More than 9%

c) Then compute it, treating $R$ as constant: the power at 200 V

W

c) By what percentage has the power fallen?

%

S7KCL at a nodePredict first

Predict first: which currents count as leaving the node?

a)$i_x$, 7 A and −2 Ab)$i_x$ and 7 A onlyc)4 A and 3 Ad)4 A, 3 A and −2 A

Then find $i_x$.

a)2 Ab)−2 Ac)6 Ad)−6 A

S8KVL around a loopPredict first

Predict first: on a clockwise walk, which elements do you cross from + to −?

a)All fourb)1 and 2 onlyc)1, 2 and 3 onlyd)1 and 3 only

Then find $v_x$.

a)−8 Vb)8 Vc)0 Vd)20 V

S9Two unknown currents

a) A KCL equation at the top node (currents leaving positive)

a)$-i_o + i_1 - 9 = 0$b)$-i_o + i_1 + 9 = 0$c)$i_o + i_1 - 9 = 0$d)$-i_o - i_1 - 9 = 0$

a) A KVL equation around the left loop (clockwise, drops positive)

a)$-60 + 4i_o + 8i_1 = 0$b)$60 + 4i_o + 8i_1 = 0$c)$-60 + 4i_o - 8i_1 = 0$d)$-60 - 4i_o + 8i_1 = 0$

b) Find $i_o$

A

b) Find $i_1$

A

c) Power of the 60 V source

W

c) The 60 V source:

a)absorbs powerb)delivers power

c) Power of the 9 A source

W

c) The 9 A source:

a)absorbs powerb)delivers power

c) Power balance: total power delivered

W

S10Which loop gives a useful KVL equation?Two-tier

In the S9 circuit, which closed path gives a KVL equation without adding a new unknown?

Answer

a)The loop through the 60 V source, the 4 Ω and the 8 Ω resistorsb)The loop through the 8 Ω resistor and the 9 A sourcec)The outer loop through the 60 V source, the 4 Ω resistor and the 9 A sourced)All three are equally useful

Reason

i)The voltage across an ideal current source is unknown; the rest of the circuit sets it.ii)The voltage across a current source is always zero.iii)KVL does not apply to a loop that contains a current source.iv)An outer loop never gives an independent equation.

S11A current-controlled current source

Find $i_\Delta$ and $v_o$, and the power of the dependent source.

Find $i_\Delta$

A

Find $v_o$

a)$v_o = 80$ Vb)$v_o = 50$ Vc)$v_o = 200$ Vd)$v_o = 100$ V

Power of the dependent source

W

S12A current-controlled voltage sourceTwo-tier

The dependent source:

Answer

a)absorbs 16 Wb)delivers 16 Wc)absorbs 8 Wd)delivers 24 W

Reason

i)$i_o$ enters its + terminal, so $p = vi > 0$.ii)Dependent sources always deliver power.iii)Its voltage is positive, so it delivers.iv)It is controlled by $i_s$, so its power is zero.

Follow-up: find $v_o$.

V

S13Spot the errorSpot the error

The worked solution below finds $v_o$. It contains the kind of mistake AI chat tools often make with dependent sources.

KVL clockwise around the left loop, drops positive: $-60 + 4i_\Delta + 12i_o = 0$.
KCL at the top node, currents leaving: $-i_\Delta + i_o + 2i_\Delta = 0$, so $i_o = -i_\Delta$.
Substitute: $-60 + 4i_\Delta - 12i_\Delta = 0$, so $i_\Delta = -7.5$ A.
Then $i_o = 7.5$ A and $v_o = 12i_o = 90$ V.
$v_o$ is positive, so the answer is correct.

b) Correct it: with the right signs, line 2 gives $i_o = k\,i_\Delta$. What is $k$?

c) Finish the solution: $v_o$

V

c) Check it with a power balance: total power delivered

W

d) Line 5 claims the answer is correct. Why is that not a check?

Answer in a sentence, then compare with the explanation.

S14Work backwards

In the S9 circuit, replace the 9 A source with a source of unknown current $I$, arrow still up. What value of $I$ makes the current in the 60 V source zero?

Answer

a)7.5 Ab)15 Ac)5 Ad)0 A

S15Modelling a car battery from measurements

A car battery is tested at three load currents:

Current drawn (A)	0	100	200
Terminal voltage (V)	12.6	11.6	10.6

Model it as an ideal voltage source in series with a resistance, as in Nilsson Example 2.9. What terminal voltage do you expect when the starter motor draws 250 A?

Answer

a)10.1 Vb)12.6 Vc)9.1 Vd)11.35 V

Follow-up: why do the headlights dim while the engine is cranking?

Answer in a sentence, then compare with the explanation.

S16Two heaters: series or parallel?Two-tierPredict first

Two identical heaters are each rated 1100 W at 220 V. You connect them in series across 220 V.

Predict first: more heat than one heater alone, or less?

a)More heat than one heater aloneb)Less heat than one heater alone

Then find the total power.

a)2200 Wb)1100 Wc)550 Wd)275 W

Reason

i)Each heater keeps $R = 44$ Ω; in series the current is $220/88 = 2.5$ A, so $P = 220 \times 2.5 = 550$ W.ii)Each heater still gets 220 V, so the powers add.iii)Two heaters always give twice the heat of one.iv)Each heater gets half the voltage, so it gives half its rated power.