Answer :
Answer:
Explanation:
primary voltage = 120 V, secondary volt = 9V
Turn ratio = turn in secondary / no of turn in primary
turn in secondary / no of turn in primary = secondary volt / primary voltage
Turn ratio = 9 / 120
= 3 / 40
Power of secondary = power of primary
Power of secondary = volt x current
= 9 x 254 x 10⁻³
= 2.286 W
power of primary ( wall socket ) = 2.286 W
Current of primary = power / volt
= 2.286 / 120
= 19.05 x 10⁻³ A .
Answer:
a) [tex]n=\frac{40}{3}[/tex]
b) [tex]I_p=0.01905\ A=19.05\times 10^{-3}\ A[/tex]
c) [tex]V_s.I_s=V_p.I_p=0.254\times 9=0.01905\times 120=2.286\ W[/tex]
Explanation:
Given:
- battery rating in voltage(output), [tex]V_s=9\ V[/tex]
- charging current used for the battery(output), [tex]I_s=254\times 10^{-3}\ A[/tex]
- voltage provided by the wall socket(input), [tex]V_p=120\ V[/tex]
a)
Turn ratio of the transformer can be given by the voltage in the input (primary) coil to the voltage at the output (secondary) coil.
[tex]n=\frac{V_p}{V_s}[/tex]
[tex]n=\frac{120}{9}[/tex]
[tex]n=\frac{40}{3}[/tex]
b)
[tex]n=\frac{I_s}{I_p}[/tex]
[tex]\frac{40}{3} =\frac{0.254}{I_p}[/tex]
[tex]I_p=0.01905\ A=19.05\times 10^{-3}\ A[/tex]
c)
The power delivered is same on the both sides of the transformer coil.
As [tex]P=V.I[/tex]
[tex]So, V_s.I_s=V_p.I_p=0.254\times 9=0.01905\times 120=2.286\ W[/tex]