Question
Find the expression (i.e., derive the expression) for the resistors connected in series.
✓
Answer
Expression for the resistance connected in series:
(i) Let $R_1, R_2$ and $R_3$ be three resistances connected in series between $C$ and $D$.
(ii) Let $R_s$ be the effective resistance in circuit and $V_1, V_2$ and $V_3$ be the potential difference across $R_1, R_2$ and $R_3$ respectively.
(iii) Let the potential difference across CD be V .
(iv) In series combination.
$v=v_1+v_2+v_3$
By using Ohm's law
$V=IR_{s} \\
\therefore V_1=IR_1, V_2=I_2 \text { and } V_3=1 R_3$
Substituting these values in equation (j) we get
$I R_S=I R_1+I R_2+I R_3 \\
\therefore R_s=R_1+R_2+R_3$
For ' $n$ ' number of resistors conneded in series we get
$R_s=R_1+R_2+R_3+\ldots \ldots \ldots \ldots \ldots \ldots \ldots+R_n$
(i) Let $R_1, R_2$ and $R_3$ be three resistances connected in series between $C$ and $D$.
(ii) Let $R_s$ be the effective resistance in circuit and $V_1, V_2$ and $V_3$ be the potential difference across $R_1, R_2$ and $R_3$ respectively.
(iii) Let the potential difference across CD be V .
(iv) In series combination.
$v=v_1+v_2+v_3$
By using Ohm's law
$V=IR_{s} \\
\therefore V_1=IR_1, V_2=I_2 \text { and } V_3=1 R_3$
Substituting these values in equation (j) we get
$I R_S=I R_1+I R_2+I R_3 \\
\therefore R_s=R_1+R_2+R_3$
For ' $n$ ' number of resistors conneded in series we get
$R_s=R_1+R_2+R_3+\ldots \ldots \ldots \ldots \ldots \ldots \ldots+R_n$
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