Consider the circuit shown in figure 32 - E8- Find the current through the 10 - resistor when the switch S is a open b closed-figure 32 - E8

R_eq = (10 + 20) Ω = 30 Ω i = ( 3/30 ) A = 0.1 A R_eq = 10 Ω i = ( 3/10 ) A = 0.3 A ...

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Standard XII

Physics

Series and Parallel Combination of Resistors

Consider the ...

Question

Consider the circuit shown in figure (32 - E8). Find the current through the 10 $Ω$ resistor when the switch S is (a) open (b) closed.

figure (32 - E8)

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Solution

$R_{e q} = (10 + 20) Ω = 30 Ω$

$i = (\frac{3}{30}) A = 0.1 A$

$R_{e q} = 10 Ω$

$i = (\frac{3}{10}) A = 0.3 A$

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Consider the circuit shown in figure (32 - E8). Find the current through the 10 Ω resistor when the switch S is (a) open (b) closed. figure (32 - E8)

Req=(10+20)Ω=30Ω i=(330)A=0.1A Req=10Ω i=(310)A=0.3A

Consider the circuit shown in figure (32 - E8). Find the current through the 10 $Ω$ resistor when the switch S is (a) open (b) closed.

figure (32 - E8)

$R_{e q} = (10 + 20) Ω = 30 Ω$

$i = (\frac{3}{30}) A = 0.1 A$

$R_{e q} = 10 Ω$

$i = (\frac{3}{10}) A = 0.3 A$