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Physics

Equivalent Resistance of a System of Resistor

Resistors R 1...

Question

Resistors $R_{1}$ =10 $Ω$ , $R_{2}$ =4 $Ω$ , $R_{3}$ = 30 $Ω$ , $R_{4}$ = 20 $Ω$ , $R_{5}$ = 60 $Ω$ and a 12 volt battery is connected as shown. Calculate:
(a) the total resistance
(b) the total current flowing in the circuit

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Solution

For $R_{1}$ and $R_{2}$ in parallel,

$\frac{1}{R_{p 1}}$ = $\frac{1}{R_{1}}$ + $\frac{1}{R_{2}}$ = $\frac{1}{10}$ + $\frac{1}{40}$ = $\frac{5}{40}$ = $\frac{1}{8}$

Or $R_{p 1}$ = 8 $Ω$

For $R_{3}$ , $R_{4}$ and $R_{5}$ in parallel,

On putting the values, we get, $R_{p 2}$ = 10 $Ω$

For $R_{p 1}$ and $R_{p 2}$ in series,

R = $R_{p 1}$ + $R_{p 2}$

R = 8 + 10 = 18 $Ω$

From Ohm’s law, R = $\frac{V}{I}$

We have, I = $\frac{V}{R}$

Putting values, we get, I = 0.67 A

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Resistors R1=10 Ω, R2=4 Ω, R3 = 30 Ω, R4 = 20 Ω, R5 = 60 Ω and a 12 volt battery is connected as shown. Calculate: (a) the total resistance (b) the total current flowing in the circuit

For R1 and R2 in parallel, 1Rp1 = 1R1 + 1R2 = 110 + 140 = 540 = 18 Or Rp1 = 8 Ω For R3, R4 and R5 in parallel, On putting the values, we get, Rp2= 10 Ω For Rp1 and Rp2 in series, R = Rp1 + Rp2 R = 8 + 10 = 18 Ω From Ohm’s law, R = VI We have, I = VR Putting values, we get, I = 0.67 A

Resistors $R_{1}$ =10 $Ω$ , $R_{2}$ =4 $Ω$ , $R_{3}$ = 30 $Ω$ , $R_{4}$ = 20 $Ω$ , $R_{5}$ = 60 $Ω$ and a 12 volt battery is connected as shown. Calculate:
(a) the total resistance
(b) the total current flowing in the circuit