Question
Explain series combination of resistors.
✓
Answer
$i.$ In series combination, resistors are connected in single electrical path. Hence, the same electric current flows through each resistor in a series combination.
$ii.$ Whereas, in series combination, the supply voltage between two resistors $R_1$ and $R_2$ is divided into $V_1$ and $V_2$ respectively.
$iii.$ According to Ohm's law,
$R _1=\frac{V_1}{I}, R _2=\frac{v_2}{I}$
Total Voltage, $V = V _1+ V _2$
$=I\left(R_1+R_2\right)$
$\therefore V =I R_s$
Thus, the equivalent resistance of the series circuit is, $R_s=R_1+R_2$
$iv.$ When a number of resistors are connected in series, the equivalent resistance is equal to the sum of individual resistances.
For $'n \ '$number of resistors,
$R _{ s }= R _1+ R _2+ R _2+\ldots \ldots \ldots \ldots \ldots+ R _{ n }=\sum_{i=1}^{i=n} R_i$
$ii.$ Whereas, in series combination, the supply voltage between two resistors $R_1$ and $R_2$ is divided into $V_1$ and $V_2$ respectively.
$iii.$ According to Ohm's law,
$R _1=\frac{V_1}{I}, R _2=\frac{v_2}{I}$
Total Voltage, $V = V _1+ V _2$
$=I\left(R_1+R_2\right)$
$\therefore V =I R_s$
Thus, the equivalent resistance of the series circuit is, $R_s=R_1+R_2$
$iv.$ When a number of resistors are connected in series, the equivalent resistance is equal to the sum of individual resistances.
For $'n \ '$number of resistors,
$R _{ s }= R _1+ R _2+ R _2+\ldots \ldots \ldots \ldots \ldots+ R _{ n }=\sum_{i=1}^{i=n} R_i$
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