Three resistors $1\ \Omega,\ 2\ \Omega$ and $3\ \Omega$ are combined in series. What is the total resistance of the combination?
If the combination is connected to a battery of emf $12 V$ and negligible internal resistance, obtain the potential drop across each resistor.
If the combination is connected to a battery of emf $12 V$ and negligible internal resistance, obtain the potential drop across each resistor.
Exercise
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Given, three resistors of resistances $1\ \Omega,\ 2\ \Omega\ \text{and}\ 3\ \Omega$ combined in series.
Therefore,
$=1+2+3$
$=6\ \Omega$
$\text{I}=\frac{\text{E}}{\text{R}+\text{r}}=\frac{12}{6+0}=2\ \text{A}$
Potential drop across $R_1 = 2 \times 1V = 2V$
Potential drop across $R_2 = 2 \times 2V = 4V$
Potential drop across $R_3 = 2 \times 3V = 6V$
Therefore,
- Total resistance of series combination is given by,
$=1+2+3$
$=6\ \Omega$
- If the combination is connected to a battery of $12 V$ and negligible internal resistance.
$\text{I}=\frac{\text{E}}{\text{R}+\text{r}}=\frac{12}{6+0}=2\ \text{A}$
Potential drop across $R_1 = 2 \times 1V = 2V$
Potential drop across $R_2 = 2 \times 2V = 4V$
Potential drop across $R_3 = 2 \times 3V = 6V$
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