I have some questions about direct current circuit questions 1. Three resistors
ID: 2013496 • Letter: I
Question
I have some questions about direct current circuit questions
1. Three resistors connected in parallel have individual values of 67.0, 6.0, and 10.0 ohms, repectively. If this combination is connected in series with a 12.0V battery and a 2.0ohm resisotr. what is the current in the 10ohm resistor (in A)? (I got 1.15A but the answer is 0.77A I coudlnt' figure out that why the current is 0.77A)
2. Four resistors, R1,R2,R3, and R4, are arrayed as follows in a wheatstone bridge circuit between four junction points, a,b,c and d: R1 is between points a and b; R2 is between b and c; R3 is between a and d; R4 is between c and d. A galvanometer and switch are connected in series between b and d. This system is connected in series to a battery at points a and c. When R3=4 ohms and the ratio R2/R4=3, the galvanometer reads zero current with the switch closed. what is thevalue of R1 (in ohms) (the answer is 12.0A)
and if under the condition of zero galvanometer current, with switch closed, the potenial difference across R4 is 3V, then what is the voltage across R2(in V)? (answer is 3.0V)
Explanation / Answer
I can help out with the first one, at least. Don't have time for more.
First, find the equivalent resistance of the resistors in parallel. (Call it Req.) Req = (1/67 + 1/6 + 1/10)^-1. (Formula for parallel resistors. Now, the resistance for the entire circuit = 2 + Req. (Treat it like two resistors in parallel.) Call this R.
The voltage drops over the 3 parallel resistors must be the same, but it's NOT 12 V b/c of that 2 ohm resistor. Voltage adds in a loop, so the voltage drop across each parallel resistor = V(battery) - V(lost to 2 ohm resistor.)
How much voltage is across the 2ohm resistor? Use Ohm's law to find the current through that resistor: I = 12 / R. Then voltage across 2ohm = (12 / R) * 2 ohms = V1.
Current through 10ohm resistor = voltage / resistance = (Vbattery - V1) / (10 ohms).