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

Question

A battery of 9 V is connected in series with resistors of 0.2 &#x3A9;, 0.3 &#x3A9;, 0.4 &#x3A9;, 0.5 &#x3A9; and 12 &#x3A9;. How much current would flow through the 12 &#x3A9; resistor?

Philoid · Accepted Answer

Here a battery of 9 V is connected in series with resistors of R₁=0.2ohm, R₂=0.4ohm, R₃=0.3ohm, R₄=0.5ohm, R₅=12ohm,

So the resultant resistance = R₁ + R₂ + R₃ + R₄ + R₅

R = 0.2+0.4+0.3+0.5+12=13.4ohm

As we know that

V= IR

Thus the current flow through 12ohm resistance will be

I = V/R

I = 9/13.4

I = 0.67amp.