Circuit/Resistors Problem

by hobo skinoski
Tags: circuit or resistors
 P: 2 Four resistors are connected to a battery as shown in the figure. The current in the battery is I, the battery emf is E, and the resistor values are R1 = R, R2 = 2R, R3 = 4R, and R4 = 3R. -Determine the potential difference across each resistor in terms of E. -Determine the current in each resistor in terms of I. -In the limit that R3 → infinity, what are the new values of the current in each resistor in terms of I, the original current in the battery? 2. Relevant equations V=IR Rtotal=R1+R2+... (Series) 1/2Rtotal=1/R1 + 1/R2 ... (Parallel) 3. The attempt at a solution I've tried a bunch of different combos, and i'm just wondering if there is some sort of way to do this.
 P: 2 hmm still trying. anyone out there?
 P: 1,133 R2 and R3 are in series...we'll call their combination Rs. R4 is parallel to Rs...we'll call their combination Rp. R1 is in series with Rp...we'll call their combination Rt. Once you find Rt, the circuit should simply be a battery in series with a resistance of Rt. From this, you could find I1, the current running through the battery and R1. I1 splits into I2 and I3 at a junction. Note that the voltage between the junctions is the same per path.
HW Helper
P: 5,341
Circuit/Resistors Problem

 Quote by hobo skinoski Four resistors are connected to a battery as shown in the figure. The current in the battery is I, the battery emf is E, and the resistor values are R1 = R, R2 = 2R, R3 = 4R, and R4 = 3R. -Determine the potential difference across each resistor in terms of E. -Determine the current in each resistor in terms of I. -In the limit that R3 → infinity, what are the new values of the current in each resistor in terms of I, the original current in the battery? 2. Relevant equations V=IR Rtotal=R1+R2+... (Series) 1/2Rtotal=1/R1 + 1/R2 ... (Parallel) 3. The attempt at a solution I've tried a bunch of different combos, and i'm just wondering if there is some sort of way to do this.
As Gear300 suggests you should first find the equivalent resistance of the entire network because that will yield the current from the E source. Then it is simply a matter of splitting the current between the legs as you go through the network.

The equivalent resistance R1234 is found by finding
R23 = R2 + R3
Then
R234 = R4 || R23
Then
R1234 = R1 + R234

Then I = E/R1234

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