Practice sheet: Chapter 4, Part 1, Node Voltages and Mesh Currents
This sheet is not marked for correctness, and AI tools are allowed. The Part 1 device-free quiz, at the start of week 7, session 1 (Tue 17 Nov), is built from twins of these questions, so make sure you can do each one on your own.
Two-tier items ask for an answer and a reason. Both must be right.
Predict first: where asked, make your prediction before you calculate.
Confidence: mark each question Sure or Unsure. It is not marked; it shows you what to revise.
Conventions: node voltages are measured from the reference node, marked with the ground symbol. In KCL, count currents leaving a node as positive. Mesh currents run clockwise; in KVL, count voltage drops as positive. Check every answer: KCL must hold at every node, and the powers must add to zero.
Powers: where a part asks for the power absorbed, use the passive sign convention: $p > 0$ means the element absorbs power; $p < 0$ means it delivers power.
Equations: steps that ask you to write equations have nothing to check online. Write them on paper, then compare with the explanation.
6Check a solution with KCL at each node, KVL around an unused loop, or a power balance S2S11S12S16
S1How many equations?Two-tier
How many equations does each method need?
Answer
Reason
There are three essential nodes: the node between the source and the first resistor joins only two elements, so it is not essential. One of the three is the reference, so the node-voltage method needs 2 equations. There are three meshes, but the current source sits in the right-hand mesh only, so that mesh current is known and the mesh-current method needs 2 equations. (b) counts the reference; (d) forgets the known mesh current; reason (iv) counts the two-element node.
S2The node-voltage method
a) Write the KCL equations at nodes 1 and 2.
Work this out on paper, then compare with the explanation.
b) Solve for $v_1$
b) … and $v_2$
c) Power absorbed by the 60 V source (negative if it delivers)
c) Power absorbed by the 3 A source (negative if it delivers)
c) Check the power balance: total power absorbed by the four resistors
a) KCL, currents leaving: node 1, $(v_1 - 60)/2 + v_1/5 + (v_1 - v_2)/10 = 0$; node 2, $(v_2 - v_1)/10 + v_2/4 - 3 = 0$ (the 3 A enters node 2). b) $v_1 = 40$ V, $v_2 = 20$ V. c) The 60 V source sends $(60 - 40)/2 = 10$ A out of its + terminal: it delivers 600 W. The 3 A source pushes 3 A up into a 20 V node: it delivers 60 W. The resistors absorb $200 + 320 + 40 + 100 = 660$ W, equal to the $600 + 60$ W delivered.
S3A source to the referenceTwo-tier
With the bottom node as the reference, how many node-voltage equations must you solve?
Answer
Reason
Follow-up: find $v_2$
Follow-up: the current the 90 V source delivers
Follow-up: why does the 30 Ω resistor not appear in your equation for $v_2$?
Answer in words, then compare with the explanation.
The 90 V source connects node 1 to the reference, so it fixes $v_1 = 90$ V and only $v_2$ is unknown: one equation. KCL at node 2: $(v_2 - 90)/10 + v_2/20 - 3 = 0$ gives $v_2 = 80$ V. The source delivers $90/30 + (90 - 80)/10 = 3 + 1 = 4$ A (360 W). The 30 Ω has 90 V across it whatever $v_2$ is, so it never enters the $v_2$ equation. Reasons (i) and (iv) are false but also point to (b): only the reason catches them.
S4A supernode
a) Draw the supernode. Write its KCL equation and its constraint equation.
Work this out on paper, then compare with the explanation.
b) Find $v_1$
b) … and $v_2$
c) Power absorbed by the 5 V source (negative if it delivers)
a) The 5 V source joins nodes 1 and 2, neither of them the reference, so one supernode encloses both nodes and the source. KCL out of the supernode: $(v_1 - 40)/4 + v_1/10 + v_2/5 - 2 = 0$, with the constraint $v_2 - v_1 = 5$. b) $v_1 = 20$ V, $v_2 = 25$ V. c) 5 A arrives at node 1 through the 4 Ω and 2 A leaves through the 10 Ω, so 3 A flows through the 5 V source from node 1 to node 2, entering its − terminal: it delivers 15 W.
S5A node equation with a dependent source
a) Write the two KCL equations and the constraint equation for $i_x$.
Work this out on paper, then compare with the explanation.
b) Find $v_1$
b) … $v_2$
b) … and $i_x$
c) Power absorbed by the dependent source (negative if it delivers)
a) $(v_1 - 72)/3 + v_1/6 + (v_1 - v_2)/2 = 0$; $(v_2 - v_1)/2 + v_2/4 - 2i_x = 0$; constraint $i_x = (v_1 - v_2)/2$. b) $v_1 = 42$ V, $v_2 = 36$ V, $i_x = 3$ A. c) The source pushes $2i_x = 6$ A up into a 36 V node: it delivers $6 \times 36 = 216$ W. The 72 V source delivers 720 W.
S6A dependent voltage source
As in Nilsson Example 4.4.
a) Write the node equations and the constraint equation.
Work this out on paper, then compare with the explanation.
b) Find $v_1$
b) … $v_2$
b) … and $i_x$
c) The current in the 4 Ω resistor, downward
c) Power absorbed by the dependent source (negative if it delivers)
a) $(v_1 - 60)/2 + v_1/10 + (v_1 - v_2)/10 = 0$; $(v_2 - v_1)/10 + v_2/20 + (v_2 - 2i_x)/4 = 0$; constraint $i_x = (v_1 - v_2)/10$. b) $v_1 = 45$ V, $v_2 = 15$ V, $i_x = 3$ A. c) The source is $2i_x = 6$ V; the 4 Ω carries $(15 - 6)/4 = 2.25$ A down into its + terminal, so it absorbs $6 \times 2.25 = 13.5$ W.
S7The mesh-current method
a) Write the two mesh equations.
Work this out on paper, then compare with the explanation.
b) Find $i_a$
b) … $i_b$
b) … and the current in the 6 Ω middle resistor, downward
c) Power absorbed by the 60 V source (negative if it delivers)
c) Power absorbed by the 10 V source (negative if it delivers)
a) $-60 + 6i_a + 6(i_a - i_b) = 0$; $2i_b + 10 + 6(i_b - i_a) = 0$. b) $i_a = 7$ A, $i_b = 4$ A; the middle 6 Ω carries $i_a - i_b = 3$ A down. c) The 60 V source delivers $60 \times 7 = 420$ W; $i_b = 4$ A enters the 10 V source’s + terminal, so it absorbs 40 W.
S8A current source in one meshPredict first
Predict first: how many mesh equations do you need?
a) Find $i_a$
a) … and $i_b$
b) The voltage across the 3 A source (+ at its top)
b) Does it absorb or deliver power? Enter the power absorbed by the 3 A source (negative if it delivers)
One equation. a) The 3 A source is in the right-hand mesh only, and clockwise $i_b$ runs down the right branch, with the source, so $i_b = 3$ A without an equation. Mesh a: $-60 + 5i_a + 10(i_a - 3) = 0$ gives $i_a = 6$ A. b) $v = 10(6 - 3) - 4(3) = 18$ V; 3 A enters the + terminal, so the source absorbs 54 W.
S9A supermesh
a) Write the supermesh KVL equation and the constraint equation.
Work this out on paper, then compare with the explanation.
b) Find $i_a$
b) … and $i_b$
c) The voltage across the 2 A source (+ at its top)
c) Power absorbed by the 2 A source (negative if it delivers)
a) The 2 A source is shared by meshes a and b, so the supermesh goes around the outside of both: $-50 + 3i_a + 2i_b + 2i_b = 0$. The source arrow points up, the direction of $i_b$ in the middle branch, so the constraint is $i_b - i_a = 2$. b) $i_a = 6$ A, $i_b = 8$ A. c) $v = 50 - 3(6) = 32$ V, + at the top; 2 A leaves the + terminal, so the source delivers 64 W.
S10A mesh equation with a dependent source
a) Write the two mesh equations and the constraint equation for $i_x$.
Work this out on paper, then compare with the explanation.
b) Find $i_a$
b) … $i_b$
b) … and $i_x$
c) Power absorbed by the dependent source (negative if it delivers)
a) $-36 + 2i_a + 5(i_a - i_b) = 0$; $3i_b + 2i_x + 5(i_b - i_a) = 0$; constraint $i_x = i_a - i_b$ (the $i_x$ arrow points down, the direction of $i_a$ in the middle branch). b) $i_a = 8$ A, $i_b = 4$ A, $i_x = 4$ A. c) The source is $2i_x = 8$ V and $i_b = 4$ A enters its + terminal: it absorbs 32 W.
S11Spot the errorSpot the error
The worked solution below finds $v$ and the power of the 60 V source. It contains the kind of direction slip AI chat tools often make.
Reference: bottom node; one unknown, $v$.
KCL at the top node, currents leaving: $(v - 60)/5 + v/10 + v/5 - 3 = 0$, so $v = 30$ V.
The current in the 5 Ω resistor from the node toward the source is $(30 - 60)/5 = -6$ A.
So 6 A flows into the source’s + terminal, and the 60 V source absorbs 360 W.
Check: $v = 30$ V is less than 60 V, so the answer is reasonable.
a) Which is the first wrong line?
b) Correct it: the power absorbed by the 60 V source (negative if it delivers)
c) Check the corrected answer with a power balance: the total power delivered
d) Line 5 calls itself a check. Why does it prove nothing?
Answer in words, then compare with the explanation.
Line 3 is right: −6 A from the node toward the source means 6 A flows from the source to the node, out of its + terminal, so the 60 V source delivers 360 W; line 4 has the direction backwards. c) Delivered: $360 + 90 = 450$ W (the 3 A source delivers $30 \times 3$); absorbed: $180 + 90 + 180 = 450$ W. d) A value between 0 and 60 V says nothing about the sign of a power; only KCL or a power balance checks it.
S12Spot the error in a mesh solutionSpot the error
Line 2 simplifies to $5i_a = -30$, so $i_a = -6$ A, and line 1 then gives $i_b = -26.8$ A.
Both currents are negative, so both meshes really circulate counterclockwise.
a) Which is the first wrong line?
b) Correct it and find $i_a$
b) … and $i_b$
c) Check with a power balance: power absorbed by the 30 V source (negative if it delivers)
c) … and the total power delivered
Line 2 is wrong: in mesh b’s own direction the shared resistor carries $i_b - i_a$, so the equation is $5i_b + 30 + 5(i_b - i_a) = 0$. Then $i_a = 10$ A, $i_b = 2$ A. Check: the 80 V source delivers 800 W; the resistors absorb $400 + 320 + 20$ W, and the 30 V source absorbs 60 W, 800 W in all.
S13Node voltages or mesh currents?
a) How many equations does the node-voltage method need?
a) How many does the mesh-current method need?
b) Solve with the method that needs fewer, and find $v$
c) The current in the 20 V source
c) Is the source being charged?
a) Two essential nodes: 1 node equation; three meshes: 3 mesh equations. b) $(v - 60)/4 + v/12 + v/6 + (v - 20)/2 = 0$ gives $v = 25$ V. c) $(25 - 20)/2 = 2.5$ A enters the 20 V source’s + terminal: it is being charged (50 W).
S14Work backwards
In the S7 circuit, replace the 10 V source with an unknown source $V$ (+ at top). What value of $V$ makes the current in the 2 Ω resistor zero?
Answer
With no current in the 2 Ω, $i_b = 0$, so $i_a = 60/12 = 5$ A and the middle 6 Ω has $6 \times 5 = 30$ V across it. The 2 Ω then has no voltage across it, so the source must match the 6 Ω: $V = 30$ V. (b) is the source voltage; (c) the original value.
S15Jump-starting a carPredict first
Each battery’s resistance includes its share of the jumper cables.
a) Before cranking, the starter is not connected. Write the node equation and find $v$
a) What current flows in the flat battery?
a) In which direction?
a) Is it being charged?
b) Predict first: while the starter cranks, does the flat battery help or keep charging?
b) Then find $v$
b) … the starter current
b) … the current in the good battery
b) … and the current in the flat battery
a) $(v - 12.6)/0.05 + (v - 12)/0.1 = 0$ gives $v = 12.4$ V. The flat battery carries $(12.4 - 12)/0.1 = 4$ A into its + terminal: it is being charged. b) Add the starter current $v/0.1$: $(v - 12.6)/0.05 + (v - 12)/0.1 + v/0.1 = 0$ gives $v = 9.3$ V. The starter draws $9.3/0.1 = 93$ A; the good battery supplies $(12.6 - 9.3)/0.05 = 66$ A; the flat battery supplies $(12 - 9.3)/0.1 = 27$ A out of its + terminal: it helps.
S16Check a solution without solving again
Two classmates report their answers. Student 1: $v_1 = 30$ V, $v_2 = 20$ V. Student 2: $v_1 = 28$ V, $v_2 = 24$ V. Without solving the circuit, use KCL at each node to decide which answer is right.