Question

A battery of 9v is connected in series with resistors of 0.2ω, 0.3 ω, 0.4 ω, 0.5 ω and 12 ω, respectively. how much current would flow through the 12 ω resistor?

Answer 1

$\huge\textbf{ANSWER}$

$<b>There is no current division occurring in a series circuit. Current flow through the component is the same.$

Given that.........

Resistances are connected in series.
Hence, the sum of the resistances will give the value of R. R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω

Potential difference, V = 9 V

$I = \: \frac{V }{R} \\ \\ I \: = \: \: \frac{9V \: }{13.4Ω \: } \\ \\ I = 0.671A \:$

∴ the current that would flow through the 12 Ω resistor will be 0.671 A.

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Answer 2

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$\bold\red{hello...frd\:swigy\:here}$

$\bold\green{Your\:Answer:--}$

Total resistance of resistors when connected in series is given by

= 1 + 2 + 3 + 4 + 5

= 0.2 Ω+ 0.3 Ω+ 0.4 Ω+ 0.5 Ω+ 12 Ω = 13.4 Ω

According to Ohm’s law,

V = IR

I =V/R

I = 9 /13.4 = 0.67

▶▶There is no current division occurring in a series circuit.

So, the current through the 12 Ω resistor will be same as 0.67 A.

I hope, this will help you___❤❤

Thank you

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