- #1
coolnufc
- 23
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Im stuck on this question.
There is a image of a network. P-Q-R
P goes to Q, and then splits into 2 routes into R.
On each route there is a resistor.
The resistor P-Q is 5 ohms, and the resistors on the 2 different routes Q-R are 3 ohms, and the other is 6 ohms.
The p.d. across PR is 14 volts.
Calculate
a) the equivalent resistance of network QR.
Wouldnt this just be 9ohms. 6+3?
b) the current flowing through PQ.
i did 14/5 = 2.8A
c) the p.d across PQ.
really confused, why is it not just 14V
d) the current through the 3 ohm resistor.
i did 7/3 = 2.3A
I used the equation R = V/I
There is a image of a network. P-Q-R
P goes to Q, and then splits into 2 routes into R.
On each route there is a resistor.
The resistor P-Q is 5 ohms, and the resistors on the 2 different routes Q-R are 3 ohms, and the other is 6 ohms.
The p.d. across PR is 14 volts.
Calculate
a) the equivalent resistance of network QR.
Wouldnt this just be 9ohms. 6+3?
b) the current flowing through PQ.
i did 14/5 = 2.8A
c) the p.d across PQ.
really confused, why is it not just 14V
d) the current through the 3 ohm resistor.
i did 7/3 = 2.3A
I used the equation R = V/I