Question

A battery of 13 V is connected in series with resistors of $2\Omega ,3\Omega ,4\Omega ,5\Omega$ and $12\Omega$, respectively. How much current would flow through the $12\Omega$ resistors?

• A. 0.5 A
• B. 1 A
• C. 5 A
• D. 7 A

Answer 1

The correct option is A

0.5 A

There is no current division occurring in a series circuit.
Current flow through each component is the same, given by Ohm’s law as V = IR
$\Rightarrow I=\frac{V}{R}$
where, R is the equivalent resistance of resistors of $2\Omega ,3\Omega ,4\Omega ,5\Omega$ and $12\Omega$.
These are connected in series.
Hence, the sum of the resistances will give the value of R.
Equivalent Resistance, $R=2+3+4+5+12=26\Omega$
Potential difference, V = 13 V
I = $\frac{13}{26}$ = 0.5 A
Therefore, the current that would flow through the $12\Omega$ resistor is 0.5 A.