A battery of $6 V$ is connected in series with resisters of $0.1$ ohm, $0.15$ ohm, $0.2$ ohm, $0.25$ ohm, and $6$ ohms. How much current would flow through the $0.3$ ohm resistor?

$0.895 A$
$2.22 A$
$1 A$
None of these

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Solution

The correct option is A
$0.895 A$

Step 1. Given data:
Voltage of battery = $6 V$
Resistors in series of = $R_{1}$ = $0.1$ ohms, $R_{2}$ = $0.15$ ohms, $R_{3}$ = $0.2$ ohms, $R_{4}$ = $0.25$ ohms, and $R_{5}$ = $6$ ohms
We have to find the current through the $0.3$ ohm resistor.
Step 2. Formula used:
We know in series combination, the total resistance $R_{t} = R_{1} + R_{2} + R_{3} + R_{4} + . . . . . . . .$
Where $R_{1}, R_{2}, R_{3}$ are resistances in series.
Also, according to Ohm's law, $V = I R$ , where $V =$ voltage, $I =$ current, $R =$ resistance,
Step 3. Calculations of current through $0.3 Ω$ resistor:
$R = R_{1} + R_{2} + R_{3} + R_{4} + R_{5}$
$R_{t} = 0.1 + 0.15 + 0.2 + 0.25 + 6 = 6.7 Ω$
Thus, the net resistance of the circuit is $6.7 Ω$
We know in series combination, the current through each and every resistor is same.
Hence the current through $0.3 Ω$ resistor is :
$V = I R I = \frac{V}{R} = \frac{6}{6.7} = 0.895 A$
So, the current through $0.3 Ω$ resistor is $0.895 A$ .
Hence, the correct option is option A.

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A battery of 6V is connected in series with resisters of 0.1 ohm, 0.15 ohm,0.2 ohm,0.25 ohm, and 6 ohms. How much current would flow through the 0.3 ohm resistor? 0.895A2.22A1ANone of these

A battery of $6 V$ is connected in series with resisters of $0.1$ ohm, $0.15$ ohm, $0.2$ ohm, $0.25$ ohm, and $6$ ohms. How much current would flow through the $0.3$ ohm resistor?

$0.895 A$
$2.22 A$
$1 A$
None of these