If the reading of ammeter $A_{1}$ in figure is 2.4 A, what will the ammeters $A_{2}$ and $A_{3}$ read? Neglect the resistance of the ammeters.

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Solution

The correct option is B 4
$V_{A B} = 20 \times 2.4$ $= 48 V$

$\frac{V_{A B}}{R} = I_{2} \Rightarrow I_{2} = \frac{48}{30} = 1.6 A$

$I_{3} = \frac{V_{A B}}{R Q},$

$R Q = \frac{30 \times 20}{50}$

$R Q = 12 Ω$

$I_{3} = \frac{48}{12} = 4 A$

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If the reading of ammeter A1 in figure is 2.4 A, what will the ammeters A2 and A3 read? Neglect the resistance of the ammeters.

The correct option is B 4VAB=20×2.4 =48V VABR=I2⇒I2=4830=1.6A I3=VABRQ, RQ=30×2050 RQ=12Ω I3=4812=4A

If the reading of ammeter $A_{1}$ in figure is 2.4 A, what will the ammeters $A_{2}$ and $A_{3}$ read? Neglect the resistance of the ammeters.

The correct option is B 4
$V_{A B} = 20 \times 2.4$ $= 48 V$

$\frac{V_{A B}}{R} = I_{2} \Rightarrow I_{2} = \frac{48}{30} = 1.6 A$

$I_{3} = \frac{V_{A B}}{R Q},$

$R Q = \frac{30 \times 20}{50}$

$R Q = 12 Ω$

$I_{3} = \frac{48}{12} = 4 A$