Solving Circuit Analysis Problems: Tips for Finding Current and Voltage Values

[gl0ck]

Hi there,
I have difficulties with the last 2 parts of the question - b) and c)

The Attempt at a Solution

I think I found the right values for I1, I2 , I3
I1 = 1A
I2 = I3 = 0.5A
Voltage drops V1 - V4 using E=I * R (16,16,8,8,16,8)
Powers - 16,4,8,4,8,8

Regards

 

[CWatters]

I think I found the right values for I1, I2 , I3
I1 = 1A
I2 = I3 = 0.5A

I agree.

Voltage drops V1 - V4 using E=I * R (16,16,8,8,16,8)
Powers - 16,4,8,4,8,8

V1 to V4 I made 16, 16, 16, 8 Volts Note V3 is across R5 not R3, V4 is across R6 not R4
P1 to P6 I made 16, 8, 4, 4, 8, 8 Watts Total = 48W.

Power from Battery = Vbat * Ibat = 48V x 1A = 48W eg same as dissipated in the R's.

b) Hint: remember that energy = power * time

c) Hint: A 60AH (AH =Amp Hour) battery can deliver 60A for one hour or pro rata.

 

[gl0ck]

Thanks for the reply!

I've just rewritten the values incorrectly..
For b) I found that for 30min it the value will be 1440 and for an hour 2880 but don't know how to end up with a good conclusion.
And still wondering for c) so these 60Ah will be enough for a little more than an hour?

 

[CWatters]

b) You made a small mistake. Time has to be in seconds not mins.

The battery delivers 1A at 48V for 30min. The power is therefore 48W.

Energy(Joules) = Power(Watts) * Time(seconds)

= 48 * 30 * 60
= 86400 Joules.

c) If the battery has a capacity of 60AH it can in theory deliver 60A for 1 hour or 1A for 60 Hours (or any other product that equals 60). If the circuit only draws 1A the battery could power it for 60 hours.

Since in part b) the circut draws 1A for 0.5 hour it has consumed only 1 * 0.5 = 0.5AH. You can work out the percentage. It's less than 1%.

 

[rezeew]

,

I understand that circuit analysis can be a challenging task, especially when it comes to finding current and voltage values. It is important to approach these problems systematically and with a clear understanding of the principles of circuit analysis.

In regards to the last two parts of the question, I would suggest using the Kirchhoff's laws to solve for the unknown values. Kirchhoff's Voltage Law states that the sum of all voltage drops in a closed loop must equal the sum of all voltage rises. This can help you determine the voltage values at different points in the circuit.

Similarly, Kirchhoff's Current Law states that the total current entering a junction must equal the total current leaving the junction. This can help you solve for the unknown current values in the circuit.

It is also important to double check your calculations and make sure they are consistent with Ohm's law, which states that the voltage drop across a resistor is equal to the product of the current flowing through it and the resistance of the resistor.

In summary, my advice for solving circuit analysis problems is to approach them systematically, use Kirchhoff's laws and Ohm's law, and double check your calculations. With practice and a good understanding of these principles, you will be able to successfully solve circuit analysis problems.