We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program


Advertise your business, and reach millions of students around the world.


All the data tables that you may search for.


For Arabic Users, find a teacher/tutor in your City or country in the Middle East.


Find the Source, Textbook, Solution Manual that you are looking for in 1 click.


Need Help? We got you covered.

Chapter 1

Q. 1.8

Consider the following circuit involving a voltage source, a current source, and two resistors.Find the voltage across the 8 Ω resistor.


Verified Solution

With the given definitions of the voltages and currents on the circuit,we need
to apply KCL and KVL systematically in order to solve the problem. One obtains
• KVL(1 → 2 → 3 → 1): v_{x}+ v_{y} − 16 = 0 \longrightarrow  v_{x}+ v_{y} = 16V,
• KCL(2): i_{x} + 3 − i_{y}= 0\longrightarrow i_{x}i_{y} = −3A.
Then, using Ohm’s law, v_{x} = 8i_{x} and v_{y} = 12i_{y}, leading to 2i_{x} + 3ii_{y}= 4. Finally, solving
for i_{x}, we obtain i_{x} = −1A, i_{y} = 2A, v_{y} = 24V, and
v_{x} = −8 V
as the voltage across the 8Ωresistor.As one verification of the solution, one can calculate
the powers of all components:
p_{8Ω} = v_{x}i_{x} = 8 W,
p_{12Ω} = v_{y}i_{y} = 48 W,
p_{16V} = 16 × (−i_{x}) = 16 W,
p_{3A} = (−v_{y}) × 3 = −72 W,
and check that
p_{8Ω} + p_{12Ω} + p_{16V}+ p_{3A} = 8 + 48 + 16 − 72 = 0.
As far as the power values are concerned, we can conclude that the current source delivers
power, while the resistors, as well as the voltage source, consume power.