Question

The resistance of the rheostat shown in the figure (32-E5) is 30 Ω. Neglecting the meter resistance, find the
figure32-E5
minimum and maximum currents through the ammeter as the rheostat is varied.

Answer 1

Answer:

Current will be minimum when the rheostat will have maximum resistance, i.e 30 Ω.

In this case, total resistance of the circuit

= [(20×10)/30]+30

= 36.67 Ω

The minimum current through the circuit,

imin = 5.5/36.67

= 0.15 A

Current will be maximum when the rheostat will have minimum resistance, i.e. 0 Ω.

In this case, total resistance of the circuit

= (20×10)/30

= 6.67 Ω

The maximum current through the circuit,

imax = 5.5/6.67

= 0.825 A

Answer 2

$i_m_i_n$= 0.15 Amp

$i_m_a_x$ = 0.825 Amp

Explanation:

Step 1:

When the rheostat have maximum resistance the current will be minimum of 30 Ω.

Circuit's total resistance = $\frac{20 \times 10}{30}+30=36.67 \Omega$

Through the circuit the minimum current is given as

$i_{\min }=\frac{5.5}{36.67}=0.15 A$

Step 2:

When the rheostat have minimum resistance the current will be maximum of 0 Ω.

In this case the circuit's total resistance is given as

$=\frac{20 \times 10}{30}=6.67 \Omega$

Maximum current through the circuit is given as

$i_{\max }=\frac{5.5}{6.67}=0.825 A$