Simple Resistor: Calculating Current

Electricity and Magnetism Level 1

The diagram above shows a simple circuit of an 11 V 11 \text{V} battery ( V V ) and two resistors ( R R ) that are in parallel followed in series by a light bulb. The resistances are as followed: R 1 = 6 Ω R_1=6 Ω , R 2 = 30 Ω R_2=30 Ω , and R 3 = 17 Ω R_3= 17 Ω ; where R 3 R_3 is the resistance of the light bulb.

Calculate the current ( I I ) of the circuit in Amps \text{Amps} .

David's Electricity Set


The answer is 0.5.

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David Hontz by David Hontz , 26, USA

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1 solution

David Hontz
Dec 28, 2016

Equivalent Resistor from Parallel Resistors: R a = ( 1 R 1 + 1 R 2 ) − 1 = ( 1 6 + 1 30 ) − 1 = ( 1 5 ) − 1 = 5 Ω R_a = \Big( \frac{1}{R_1} + \frac{1}{R_2} \Big)^{-1} = \Big( \frac{1}{6} + \frac{1}{30} \Big)^{-1} = \Big( \frac{1}{5} \Big)^{-1} = 5Ω Equivalent Resistor from Resistors in Series: R c i r c u i t = R a + R 3 = 5 + 17 = 22 Ω R_{circuit} = R_a + R_3 = 5 + 17 = 22Ω Current Calculation: I = V R c i r c u i t = 11 V 22 Ω = 1 2 A = 0.5 A I = \frac{V}{R_{circuit}} = \frac{11V}{22Ω} = \frac{1}{2}A = \boxed{0.5 A}

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Current is quantified in terms of amperes. Capacitance is quantified in terms of Farads.

Steven Chase - 4 years, 5 months ago

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Correct and thank you.

David Hontz - 4 years, 5 months ago

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