Notifications
Clear all
0
26/09/2021 6:18 pm
Topic starter
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
1 Answer
0
26/09/2021 6:20 pm
Given: V = 9 volts
Applying the formula
R = R1 + R2 + R3 + R4 + Rs, we get
R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω
By Ohm’s law, we have
V = RI
or 9 = 13.4 x I
or I = \(\frac{9}{13.4}\) = 0.67A
Hence, the current flows through the 12 Ω resistor is 0.67 A.