A battery with a terminal voltage of 12V is connected to a circuit consisting
of four 10 Ω and one 5 Ω resistor all in series. Assume the battery has
negligible internal resistance. Calculate the current through each resistor.
Answer
Equivalent resistance is
"R=4*10+5=45\\Omega"
These resistances are in series so current will be same.
"I=\\frac{V}{R}\\\\=\\frac{12}{45}\\\\=0.27A"
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