current in a circuit?

Question

A 19.9 Volt battery pushes a current around a circuit with a resistance of 33.4 Ohms. Find the current in the circuit.

Find the power dissipated in the circuit.

Answer 1

The current, I = V/R = 19.9/33.4=0.596A
Power dissipated = (I^2)R= (0.596^2)(33.4)=11.86Watt

Answer 2

For all such questions, keep in front of you these few simple relations between voltage V, resistance R, current I and power W.

I = V / R , W = I^2R or V^2/R

Now let us do the problem.

Given V = 19.9 and R = 33.4, we need to find I and W.

I = V / R = 19.9 / 33.4 = 0.5958... = 0.6 amps approx.

Power dissiplated in the circuit W is

W = V^2 / R = 19.9 X 19.9 / 33.4 = 11.8565.. = 11.9 watts

Answer 3

If the meter can handle a contemporary of V / (R1 +R2) it will be high-quality, however the measured value is purely no longer the present in the circuit whilst the meter isn't there. think of of the ammeter as a quick circuit (or almost so) as to its outcomes on the circuit. for this reason it quite is expounded in sequence. There are some contemporary meters that clamp and iron center around the twine and degree the present in AC circuits with out breaking any connections. There are some quite costly ones that do an identical for DC with extremely some electronics in them.

Answer 4

I = V/R. so the current is 19.9/33.4, but i don't know the power dissipated. sorry.

Answer 5

I = E/R = 19.9/33.4. For the power, use P = EI.

Answer 6

To find out actual current internal resistance of battery is necessary as I = V/(R+r).
But r has not been given.

Answer 7

Current=19.9V/33.4ohm
Answer=0.6A

Power= Current^2 x Resistance or V x I(for dc)
= 0.6^2x33.4 or 19.9x0.6

Answer=12W

Note: decimal points rounded; 'x' means multiplied by

Answer 8

For power it is
p=IV
p=(.596)(19.9)=11.9 W

Answer 9

i=19.9/33.4
P=19.9^2/33.4