Question

Calculate the electric current passing through the circuit given ,when key is closed​

Answer 1

Explanation:

the connection is series connection

5ohm + 10ohm +15ohm

== 30 ohm

ohm law

V=IR

V/R = I

10/30 = I

3 ampier

Answer 2

Explanation:

We have,,

Resistor-1, $\sf{R_{1}=5\:Ω}$

Resistor-2, $\sf{R_{2}=10\:Ω}$

Resistor-3, $\sf{R_{3}=15\:Ω}$

Potential Diff., $\sf{V=10\:v}$

Since, all resistors are connected in series,

Therefore,

Equivalent resistant, $\sf{R=R_{1}+R_{2}+R_{2}}$

$\implies{\sf{R=5\:Ω+10\:Ω+15\:Ω}}$

$\implies{\boxed{\sf{R=30\:Ω}}}$

By ohm's law,

Electric current passing through the circuit is given by,

$\sf{I=\frac{V}{R}}$

$\implies{\sf{I=\frac{10}{30}}}$

$\implies{\sf{I=\frac{1}{3}}}$

$\implies{\boxed{\sf{I=0.33 \:A}}}$

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