A circuit is constructed with six resistors and two batteries as shown. The batt
ID: 1323478 • Letter: A
Question
A circuit is constructed with six resistors and two batteries as shown. The battery voltages are V1 = 18 V and V2 = 12 V. The positive terminals are indicated with a + sign, The values for the resistors are: R1 = R5 = 72 ?, R2 = R6 = 116 ? R3 = 61 ?, and R4 = 88 ?. The positive directions for the currents I1, I2 and I3 are indicated by the directions of the arrows.
What is V4, the magnitude of the voltage across the resistor R4?
What is I3, the current that flows through the resistor R3? A positive value for the current is defined to be in the direction of the arrow.
What is I2, the current that flows through the resistor R2? A positive value for the current is defined to be in the direction of the arrow.
What is I1, the current that flows through the resistor R1? A positive value for the current is defined to be in the direction of the arrow.
What is V(a) V(b), the potential difference between the points a and b?
A circuit is constructed with six resistors and two batteries as shown. The battery voltages are V1 = 18 V and V2 = 12 V. The positive terminals are indicated with a + sign, The values for the resistors are: R1 = R5 = 72 ?, R2 = R6 = 116 ? R3 = 61 ?, and R4 = 88 ?. The positive directions for the currents I1, I2 and I3 are indicated by the directions of the arrows. What is V4, the magnitude of the voltage across the resistor R4? What is I3, the current that flows through the resistor R3? A positive value for the current is defined to be in the direction of the arrow. What is I2, the current that flows through the resistor R2? A positive value for the current is defined to be in the direction of the arrow. What is I1, the current that flows through the resistor R1? A positive value for the current is defined to be in the direction of the arrow. What is V(a) V(b), the potential difference between the points a and b?Explanation / Answer
but if you rearange the circuit its easy
......|---6---4----|
......|................|-----|
|----|----15-------|.....|
6............................|
|.............................|
B............................|
|-------------------------|
easier to see now
we figure out the parallel circuit first
thats Rp = 10*15/(10+15)= 150/25 = 6ohms
so ITS equivalent to 24 volts plus 2x 6 ohm resistors.
I used kirchoffs VOLTAGE law which states that sum of the voltage drops in a closed loop will be 0 .
+24 volts from batter - equal voltage across two identical resisters means THE VOLTAGES MUST BE IDENTICAL SO 24/2 = 12 VOLTS ACROSS each resister.
That means 12 volts across R1 = 6 ohm
so A) is I = 12/6 = 2 amps
you could use total resistance if you like
I = Vt/Rt= 24/12 = 2 amps.. since series circuit the 6 ohm resistor had 2 amps in it.
B) we found that to find the current so its 12 volts because the parallel circuit was 6 ohms equivalent .. so the voltage had to be spread over both equally...
If you know 2 amps.. and you know R = 6
then V = I*R = 2*6 = 12. volts drop across 6 ohm resistor
C) We know the voltage across the parallel circuit is 12 volts as stated above.
so
the current throught he 15 ohm resister is
I = V/R = 12/15 = 0.8 amps
D) as stated before if the potential difference across the EQUIVALENT resister for the parallel circuit is 12 volts.. the voltage across the parallel circuit is 12 volts.
and since 15 ohm resister connects from start to end of the parallel circuit it must be 12 volts drop.
E) since the current going in the parallel circuit is 2 amps then it must split between the branches and that means 1.2 amps goes into the 6 + 4 ohm resister combination
because 0.8 amps went through the 15 ohm resister.
The sum of I's going into a node must equal the sum of I's leaving a node.
Thats kirchoffs current law.
It = I15 + I(6+4)
2.0 = 0.8 + I(6+4)
1.2 = I (6+4) the six and 4 are the resister names
only not values to be multiplied.
The R3 = 6 ohms has 1.2 amps going through it.
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