What is the current through each resistor in this circuit?

[Rodgerd]

Consider the circuit shown in the figure . Suppose the four resistors in this circuit have the values R1 = 12 , R2 = 6.0 , R3 = 7.3 , and R4 = 15 , and that the emf of the battery is E = 18V .

(1) Find the current through each resistor using the rules for series and parallel resistors. Answer in terms of (I1, I2, I3, I4)

(2) Find the current through each resistor using Kirchhoff's rules. Answer in terms of (I1, I2, I3, I4)

 

[phinds]

Be a lot easier to help you if you actually posted the figure you are talking about.

Also, use the template and show some attempt on you own. We don't just feed answers here.

 

[Rodgerd]

Here is the image "Phinds"http://session.masteringphysics.com/problemAsset/1126221/4/5416321074_61.jpg

 

[BvU]

Now what eqns have you got at your disposal and what did you do with them so far ?

And: welcome to PF :-)

 

[HalfLostAndSearching]

(1) Using the rules for series and parallel resistors, we can calculate the total resistance of the circuit as R = R1 + R2 || R3 + R4, where || represents the parallel combination of resistors. This gives us R = 12 + (6.0*7.3)/(6.0+7.3) + 15 = 12 + 4.24 + 15 = 31.24 ohms.

From Ohm's law, we can calculate the total current in the circuit as I = E/R = 18/31.24 = 0.576 A.

Using the current divider rule, we can calculate the current through each resistor as follows:

I1 = I * (R2 || R3 + R4)/(R1 + R2 || R3 + R4) = 0.576 * (6.0*7.3)/(12 + 6.0*7.3) = 0.17 A

I2 = I * R1/(R1 + R2 || R3 + R4) = 0.576 * 12/(12 + 6.0*7.3) = 0.22 A

I3 = I * R1/(R1 + R2 || R3 + R4) = 0.576 * 7.3/(12 + 6.0*7.3) = 0.13 A

I4 = I * R4/(R1 + R2 || R3 + R4) = 0.576 * 15/(12 + 6.0*7.3) = 0.26 A

Therefore, the current through each resistor is: I1 = 0.17 A, I2 = 0.22 A, I3 = 0.13 A, I4 = 0.26 A.

(2) Using Kirchhoff's rules, we can set up the following equations:

- Loop 1: -E + R1*I1 + (R2 + R3)*I2 = 0
- Loop 2: -E + R4*I4 + (R2 + R3)*I2 = 0
- Junction rule: I1 + I2 = I3 + I4

Solving these equations simultaneously, we get:

I1 = 0.17 A, I2