# If r1 is in series with parallel connected to r2 and r3, what happens to total current if r2 opens?

Jun 16, 2015

If r1 is in series with parallel connected to r2 and r3, the total current decreases if r2 opens.

#### Explanation:

Before opening of ${r}_{2}$

${R}_{e q 1} = {r}_{1} + \frac{{r}_{2} {r}_{3}}{{r}_{2} + {r}_{3}}$

${R}_{e q 1} = {r}_{1} + \frac{{r}_{3}}{1 + {r}_{3} / {r}_{2}}$$\text{ } \textcolor{g r e e n}{\left(1\right)}$

After opening of ${r}_{2}$:

hence ${r}_{2} =$infinity$\text{ } \textcolor{g r e e n}{\left(2\right)}$

write down$\text{ } \textcolor{g r e e n}{\left(1\right)}$ after substitution of$\text{ } \textcolor{g r e e n}{\left(2\right)}$

${R}_{e q 2} = {r}_{1} + {r}_{3}$[since${r}_{3} / {r}_{2} = 0$ if ${r}_{2}$=infinity]

${R}_{e q 2} = {r}_{1} + {r}_{3}$$\text{ } \textcolor{g r e e n}{\left(3\right)}$

from $\text{ } \textcolor{g r e e n}{\left(1\right)}$&$\textcolor{g r e e n}{\left(3\right)}$

${R}_{e q 2} > {R}_{e q 1}$

If resistance value is increased under constant application voltage and constant temperature ,then the rate of change of charge with respect to time decreases according to ohm's law.
$I \propto \frac{1}{R}$ at constant voltage.

Therefore,the total current $\textcolor{b l u e}{\mathrm{de} c r e a s e s}$ if r2 opens.