117th Problem 2016

Electricity and Magnetism Level 2

In the above circuit, R 1 = 100 Ω { R }_{ 1 }=100 \ \Omega , R 2 = 300 Ω { R }_{ 2 }=300 \ \Omega , R 3 = 600 Ω { R }_{ 3 }=600 \ \Omega and V T = 150 V { V }_{ T }=150 \ V , what is the total current I T I_T of the circuit?

Part One .
Check out the set: 2016 Problems .
This is part five of a problem.


The answer is 2.25.

Angela Fajardo by Angela Fajardo , 19, Philippines

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3 solutions

Ashish Menon
Jul 15, 2016

The total resistance is ( 1 100 + 1 300 + 1 600 ) − 1 = 200 3 {\left(\dfrac{1}{100} + \dfrac{1}{300} + \dfrac{1}{600}\right)}^{-1} = \dfrac{200}{3} .
Now, using Ohm's Law:-
I = V R = 150 × 3 200 = 2.25 A I = \dfrac{V}{R} = \dfrac{150 × 3}{200} = \color{#3D99F6}{\boxed{2.25 A}} .

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Chew-Seong Cheong
Jul 15, 2016

The equivalent resistance of the circuit R e q = 1 1 R 1 + 1 R 2 + 1 R 3 = 1 1 100 + 1 300 + 1 600 = 200 3 Ω R_{eq} = \dfrac 1{\frac 1{R_1} + \frac 1{R_2} + \frac 1{R_3}} = \dfrac 1{\frac 1{100} + \frac 1{300} + \frac 1{600}} = \dfrac {200}3 \ \Omega

Therefore, the total current I T = R e q V T = 150 × 3 200 = 2.25 A I_T = \dfrac {R_{eq}}{V_T} = 150 \times \dfrac 3{200} = \boxed{2.25} \ A

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Viki Zeta
Jul 15, 2016

Refer this solution first.

Therefore, total resistance is 200 3 Ω \frac{200}{3} \Omega .

Therefore,

V = I R ⟹ I = V R = 150 200 3 = 2.25 V = IR \\ \implies I = \frac{V}{R} = \dfrac{150}{\frac{200}{3}} = 2.25

2 Helpful 0 Interesting 0 Brilliant 1 Confused

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