# Physics Question?

1) What is the current through the 24 ohm? The voltage?

2) Using kirchoff's voltage low and current law determine the voltage and current through the remaining 1 ohm resistor

Okay, so I know the equivalent resistance which is 10 ohms, total current leaving the battery which is 1.0A and the voltage across 3.0 ohm resistor which is 3.0 volts. The potential difference between the ends of the 8 ohm resistor is 6v and the current is .75A. Relevance

I agree with you. The total resistance is 10 Ω and the total current is 1 ampere. This current flows through the 3 Ω and 1 Ω resistors

V = I * R = 1 * 4 = 4 volts

10 – 4 = 6 volts

Since the 24 Ω and the 8 Ω resistors resisters are in parallel, 6 volts is the voltage for both of them. Use the following equation to determine the current for each of them.

I = V ÷ R

I 24 = 6 ÷ 24 = ¼ ampere

I 8 = 6 ÷ 24 = ¾ ampere

• You seem to have done all the hard stuff already. Perhaps it is just visualising the circuit that is the problem, I have arranged it to look more intuitive below.

Given you already know what you said, if the total current is 1A and the 8R takes 0.75A then the rest goes through the 24R, ie 0.25A.

Similarly, you know the voltage drop of 3V and 6V, so the other drop is 1V to give a total of 10V • for parallels:

1/R = 1/24 + 1/8

1/r = 1/24 + 3/24

1/r = 4/24

r = 6

So the circuit is like we have 3 resistors in series

3.0, 6.0 and 1.0

Effectively it's 3+6+1 = 10 ohms

Current from battery is

e = iR

10 = i * 10

i = 1 amp (thru 3.0 7 1.0 ohm resistors)

There are multiple techniques to get i thru 24 ohm

10v - 3*1 - 1*1 = 6 volts across 24 ohm

e = ir

i = e/r

i = 6.0v/24 ohm

i = 1/4 a