Question

9. A battery of 9 V is connected in series with resistors of 0.2 ohm,0.3 ohm, 0.4ohm,0.5ohm
and 12 ohm, respectively. How much current would flow through the 12 ohm of resistor?
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Answer 1

Answer:

0.67

Step-by-step explanation:

series me current same hota hai

total resistance =13.5

using V=IR

I=0.67

Answer 2

Answer:

0.67 A

Step-by-step explanation:

Given:-

• Battery=9V
• Resistors of 0.2 ohm, 0.3ohm, 0.4ohm, 0.5ohm and 12ohm

As all are connected in series,

The resistance would be:-

$\tt \: R_s=R_1+R_2+R_3+R_4+R_5$

=0.2+0.3+0.4+0.5+12

=13.4 Ohm

By ohms law : V=IR

I=V/R

=9/13.4

=0.67 A

As resistors are connected in series.

$\tt \: Current=I_1=I_2=I_3=I_4=I_5$

Current= 0.67 A

Hence current across 12ohms resistor is 0.67A.