can someone please please help me with these phyiscs questions?
first, what resistance must be placed in parallel with a 155 resistor to make the equivalent resistance 115 ohms?
secondly, the resistors 42 and 64 ohms, are connected in parallel. the current through the 64 ohms resistor is 3 amps. A.) determine the current in the other resistors B.) what is the total power consumed by the two resistors?
and lastly, a coffee cup heater and a lamp are connected in parallel to the same 120
V outlet. Together, they use a total of 84 watts of power. The resistance of the heater is 6.0X10^2 ohms. Find the resistance of the lamp.
please any help is appreciated.
first, what resistance must be placed in parallel with a 155 resistor to make the equivalent resistance 115 ohms?
secondly, the resistors 42 and 64 ohms, are connected in parallel. the current through the 64 ohms resistor is 3 amps. A.) determine the current in the other resistors B.) what is the total power consumed by the two resistors?
and lastly, a coffee cup heater and a lamp are connected in parallel to the same 120
V outlet. Together, they use a total of 84 watts of power. The resistance of the heater is 6.0X10^2 ohms. Find the resistance of the lamp.
please any help is appreciated.
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1) By 1/R = 1/R1 + 1/R2
=>1/115 = 1/155 + 1/R
=>1/R = 1/115 - 1/155
=>R = 445.625Ω
2)(A) By 1/R = 1/R1 + 1/R2
=>1/R = 1/42 + 1/64
=>R = 25.36
=>By V = i x R
=>V = i2 x R2
=>V = 3 x 64 = 192 V
=>By V = i1 x R1
=>192 = i1 x 42
=>i1 = 4.57 amp
(B) By P = V^2/R
=>P = (192)^2/25.36
=>P = 1453.63 Watt
3) By P = V^2/R
=>84 = (120)^2/R
=>R = 171.43Ω
=>By 1/R = 1/R1 + 1/R2
=>1/171.43 = 1/600 + 1/R
=>1/R = 1/171.43 - 1/600
=>R = 240Ω
=>1/115 = 1/155 + 1/R
=>1/R = 1/115 - 1/155
=>R = 445.625Ω
2)(A) By 1/R = 1/R1 + 1/R2
=>1/R = 1/42 + 1/64
=>R = 25.36
=>By V = i x R
=>V = i2 x R2
=>V = 3 x 64 = 192 V
=>By V = i1 x R1
=>192 = i1 x 42
=>i1 = 4.57 amp
(B) By P = V^2/R
=>P = (192)^2/25.36
=>P = 1453.63 Watt
3) By P = V^2/R
=>84 = (120)^2/R
=>R = 171.43Ω
=>By 1/R = 1/R1 + 1/R2
=>1/171.43 = 1/600 + 1/R
=>1/R = 1/171.43 - 1/600
=>R = 240Ω