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

figure (32-E5)

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Solution

$i_{m i n} = \frac{5.5 \times 3}{110} = \frac{(1.5)}{10} = 0.15$

$i_{m a x} = \frac{5.5 \times 3}{20} = \frac{8.25}{10} = 0.825$

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

imin=5.5×3110=(1.5)10=0.15 imax=5.5×320=8.2510=0.825

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

figure (32-E5)

$i_{m i n} = \frac{5.5 \times 3}{110} = \frac{(1.5)}{10} = 0.15$

$i_{m a x} = \frac{5.5 \times 3}{20} = \frac{8.25}{10} = 0.825$