Question 178: Current Electricity

Question

Refer to the figure given below. The values of $I_1, I_2$ and $I_3$ are $\_\_\_\_$ .

Accepted Answer

For a voltmeter, to increase its range, we connect a
high resistance in series
. Let us calculate it.
The voltmeter has internal resistance $x \, \Omega$ and can measure up to $20 \, 	ext{V}$.
So, the full-scale current of the voltmeter is
$$ I = \frac{V}{R} = \frac{20}{x} $$
Now we want the new range to be $30 \, 	ext{V}$.
Let the additional resistance connected in series be $R$.
Then for the same full-scale current,
$$ 30 = I(x + R) $$
Substitute $I = \frac{20}{x}$:
$$ 30 = \frac{20}{x}(x + R) $$
$$ 30x = 20(x + R) $$
$$ 30x = 20x + 20R $$
$$ 10x = 20R $$
$$ R = \frac{x}{2} $$
So, we must
connect a resistor of $\frac{x}{2}\,\Omega$ in series with the voltmeter
.
Therefore, the correct option is:
$$ \boxed{	ext{Option A}} $$

Answer

$I_1=2.5 \mathrm{~A}, I_2=1.875 \mathrm{~A}, I_3=1.875 \mathrm{~A}$

Answer

$I_1=1.875 \mathrm{~A}, I_2=2.5 \mathrm{~A}, I_3=1.875 \mathrm{~A}$

Answer

$I_1=1.875 \mathrm{~A}, I_2=1.875 \mathrm{~A}, I_3=2.5 \mathrm{~A}$

Answer

$I_1=2.5 \mathrm{~A}, I_2=2.5 \mathrm{~A}, I_3=1.875 \mathrm{~A}$

Current Electricity

Refer to the figure given below. The values of I_1, I_2 and I_3 are \_\_\_\_ .

Choose the correct answer: