Resistors And Battery Diagram Poser

[kuiperbelt]

The son of our next door neighbour is doing HND in Electrical and Electronic Engineering and requires some assistance.
He has asked me to help but it's a little beyond me. I told him that I'm sure someone on AB can give him a pointer if not how to arrive at the answer :-)
The question is to find the current through the 5 ohm resistor and the voltage across the top of the 2 ohm resistor and the bottom of the 20 V battery. He does have the answers but needs a method to solve the problem!! See link below:

https://ibb.co/ggqzkhF

[kuiperbelt]

Guys, Sorry to be pushy over this. Is it possible to have an explanation of your answers before 7pm tonight. Many Thanx :-)

[kuiperbelt]

Hymie >13.18

Thanx 4 your contribution but my neighbour has already done a very similar problem as part of the same classwork exercise.

The question posed in this thread is particularly tricky because there are 3 voltage sources (2 batteries and the 25v d.c supply). Anymore suggestions are very welcome!! :-)

[Etch]

OK, a brief description of my method:
Assume that the voltage at the bottom of the circuit is 0 V and the voltage at the node between the 3Ω, 2Ω and 5Ω resistors is X Volts. The voltage across the 3Ω resistor is (25 - X) volts so the current through this section, call it I1, is (25 - X)/3 Amps. This current then splits between the 2 legs containing the batteries. The voltage across the 2Ω resistor is (X - 20) Volts and therefore the current in this section, call it I2, is (X - 20)/2 Amps. The current in the 5Ω section, call it I3, is given by a similar formula. Substitute these formulae in the relationship I1 = I2 + I3 and solve for X, which gives the voltage across the 2Ω resistor and 20 V battery. Use the value of X in the formula for I3 to obtain the current through the 5Ω resistor.

[ZebuSanctuary]

Firstly look upon the 25 V d.c supply as a battery charger trying to charge the 2 batteries.

Let current through 3Ω resistor be the sum of the currents i2 and i3 which flow in the 20 V battery and 15 V battery branches respectively.

Note, Kirchhoff's loops are clockwise, voltage drops and EMFs will be treated as negative if in the clockwise direction. Current in 3Ω,
to use Hymie's terminology is Eastbound and branch currents i2 and i3 are Southbound.

Apply Kirchhoff's second law to loop 1 - comprising 25 V d.c, 3Ω , 2Ω and 20 V battery.

Then 0 = - 25 + 3(i2 + i3) + 2.i2 + 20 ---> 25 = 3(i2 + i3) + 2.i2 + 20 Rearranging this becomes ----> 5 = 5.i2 + 3.i3 ---- Eq 1

Now Kirchhoff's 2nd law to loop 2 - Comprising the 20 V battery, 2Ω, 5Ω , 15 V battery and 1Ω

Then 0 = - 20 - 2.i2 + 5.i3 + 15 + 1.i3 -----> 20 = - 2.i2 + 6.i3 + 15
Rearranging this becomes -----> 5 = -2.i2 + 6.i3 ---- Eq 2

The HND student should now identify Eq 1 & Eq 2 form a pair of simultaneous equations.

I'll leave him to solve for i2 and i3

Hope this helps :0)

[kuiperbelt]

Etch, Zebu, My student neighbour is most grateful for your contributions. He says both accounts are accurate and valid!!

He has opted to take the Kirchhoff solution as the lecturer was expecting an answer utilizing either Kirchhoff, Superposition or Thevenins. He will not get the results until next week!!

From me, many thanks to the both of you!!