Four resistors are connected to a battery as shown in thefigure. The current through the battery is I, the battery'selectromotive force (emf) is & = 5.75 V, and the resistorvalues are R₁ = R, R₂ = 2R, R3 = 4R, and R₁ = 3R. Findthe voltages across each resistor.V₁ =V/₂ =V3 =V4 =VVVVR₁ = RwwwR₂ = = 2RwwwR₁=3RwwR₂ = 4R

From the given diagram we can say that and are in series, by adding them we get equivalent…

Answered: Four resistors are connected to a…

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter29: Direct Current (dc) Circuits

Section: Chapter Questions

Problem 64PQ: Ralph has three resistors, R1, R2, and R3, connected in series. When connected to an ideal emf...

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Question

Four resistors are connected to a battery as shown in the
figure. The current through the battery is I, the battery's
electromotive force (emf) is & = 5.75 V, and the resistor
values are R₁ = R, R₂ = 2R, R3 = 4R, and R₁ = 3R. Find
the voltages across each resistor.
V₁ =
V/₂ =
V3 =
V4 =
V
V
V
V
R₁ = R
www
R₂ = = 2R
www
R₁=3R
ww
R₂ = 4R

Expert Solution

Step 1: Calculation of the equivalent resistance of the circuit

From the given diagram we can say that $R subscript 2$ and $R subscript 3$ are in series, by adding them we get equivalent resistance in series is

$R subscript s equals R subscript 2 plus R subscript 3 R subscript s equals 2 R plus 4 R R subscript s equals 6 R$

Now this $R subscript s$ is in parallel with resistance $R subscript 4$ , then equivalent resistance in parallel is

$1 over R subscript p equals 1 over R subscript s plus 1 over R subscript 4 rightwards double arrow 1 over R subscript p equals fraction numerator 1 over denominator 6 R end fraction plus fraction numerator 1 over denominator 3 R end fraction rightwards double arrow R subscript p equals fraction numerator 3 R cross times 6 R over denominator 3 R plus 6 R end fraction rightwards double arrow R subscript p equals 2 R$

Now this $R subscript p$ is in series with the resistance $R subscript 1$ then equivalent resistance for the circuit is,

$R subscript e q end subscript equals R subscript p plus R subscript 1 R subscript e q end subscript equals 2 R plus R R subscript e q end subscript equals 3 R$

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