In the given (a) calculate the effective resistance of the circuit and (b) calculate the reading of the ammeters $A_{1} A_{2}$ and $A_{1}$ If $R_{1} = 10 Ω$ and $R_{2} = 6 Ω$ and $E = 10 V$

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Solution

$R_{1} = 10$
$R_{2} = 6$
$(1)$ $R_{n e t} = 24.831 Ω$
$I = \frac{E}{R} = \frac{10}{24.831} = 0.4027$
$A_{1} = 0.4027 A$
$A_{2} = 0.2013 A$
$A_{3} = \frac{24.82}{30.82} \times 0.2013$
$A_{3} = 0.1621 A$

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In the given (a) calculate the effective resistance of the circuit and (b) calculate the reading of the ammeters A1A2 and A1 If R1=10Ω and R2=6Ω and E=10V

R1=10R2=6(1) Rnet=24.831ΩI=ER=1024.831=0.4027A1=0.4027AA2=0.2013AA3=24.8230.82×0.2013A3=0.1621A

In the given (a) calculate the effective resistance of the circuit and (b) calculate the reading of the ammeters $A_{1} A_{2}$ and $A_{1}$ If $R_{1} = 10 Ω$ and $R_{2} = 6 Ω$ and $E = 10 V$

$R_{1} = 10$
$R_{2} = 6$
$(1)$ $R_{n e t} = 24.831 Ω$
$I = \frac{E}{R} = \frac{10}{24.831} = 0.4027$
$A_{1} = 0.4027 A$
$A_{2} = 0.2013 A$
$A_{3} = \frac{24.82}{30.82} \times 0.2013$
$A_{3} = 0.1621 A$