R = 14/8.
Add 11 in series = 12.75 Ohm.
If NOW 3 amps flows through complete circuit of 12.75 Ohms, voltage required would be:-
V = I x R
= 3 x 12.75
= 38.25 v, or 38 to whole numbers, As demanded!
two resistors in parallel 1/Rt = 1/R1 +1/R2
let R1 =14 and R2 =2 then Rt = 1.75 ohms
Series resistors add so 1.75 + 11 = 12.75 ohms
total current = 3Amps
V = IR then 3 ( 12.75) = 38.25≈ 38V
V total = (Amp total)*(Resistance total)
V total = (3A)*[(14 ll 2) + (11) Ohm] = (3A)*[(1.75 + 11)Ohm] = (5.25 + 33) Volts ~ 38 Volts
5.25V
rounds to 5 V
It's very simple Ohm's Law. Did you not pay any attention at all in class?
38volts.
A circuit consiting of two resistors of 14 and 2 Ohms, joined in parallel is connected in series with a third resistance of 11 Ohms. A current of 3 A flows through the circuit
Calculate the total voltage in the circuit
Answer to the nearest whole number
