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Equivalent Resistance of a System of Resistor

Two resistors...

Question

Two resistors of $2.0 Ω$ and $3.0 Ω$ are connected (a) in series, (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case, draw a circuit diagram and calculate the current through the battery.

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Solution

(a) In series

$R_{e q} = R_{1} + R_{2} = 2 + 3 = 5 Ω$

Current, $I = \frac{V}{R_{e q}} = \frac{6}{5} = 1.2 A$

(b) In parallel:

$R subscript 1 space end subscript space a n d space R subscript 2 space a r a e space c o n n e t e d space i n space p a r a l l e l 1 over R equals 1 over R subscript 1 plus 1 over R subscript 2 equals 1 half plus 1 third equals 5 over 6 R equals 1.2 space o h m V equals 6 V I equals V over R equals fraction numerator 6 over denominator 1.2 end fraction equals 5 A$

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Equivalent Resistance of a System of Resistor

Standard X Physics

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Two resistors of 2.0Ω and 3.0Ω are connected (a) in series, (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case, draw a circuit diagram and calculate the current through the battery.

(a) In series Req=R1+R2=2+3=5 Ω Current, I=VReq=65=1.2 A (b) In parallel:

Two resistors of $2.0 Ω$ and $3.0 Ω$ are connected (a) in series, (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case, draw a circuit diagram and calculate the current through the battery.