On the left side is a 120V battery. On the top there is a resister of 2 ohms and then a battery of 12. On the right side there is a resisor of 4 ohms. On the bottom there is a resistor of 4 followed by a battery of 12V and then a resistor R. What must the value of R be if the circuit current is to be 1.0 A
2007-03-19 05:13:33 · 4 answers · asked by theman 1 in Science & Mathematics ➔ Physics
what would be the thermal energy that appears in that problem
2007-03-19 05:28:01 · update #1
I am totally confused.
2007-03-19 05:35:08 · answer #1 · answered by confused 3 · 0⤊ 0⤋
The method of Justin H is OK but he assumed some things you didn't say. Is it really 120V?
Are all 3 batteries in the same polarity? Meaning, if you make a clockwise loop around the circuit, when you get to each battery, will you always encounter the - terminal first and then the +? If so, add the voltages and proceed like Justin H did. Otherwise subtract any battery voltage if the orientation is opposite.
2007-03-19 12:04:28 · answer #2 · answered by sojsail 7 · 0⤊ 0⤋
assumeing it's not a 120V battery! the answer to that would be fire...
add the batteries add the resistors since it's a series ciruict
V=IR
or
R=V/I
Battery total = V= 12 +12 +12 =36
I = 1.A (given)
R = 36
then subtract the known R values
36- 2 - 4 - 4 = 26
R =26 for the unknown
2007-03-19 05:23:03 · answer #3 · answered by Justin H 4 · 0⤊ 0⤋
Top, Bottom, Left, and Right are not words to describe a circuit. Draw a picture or explain your circuit in terms of series, or parallel.
2007-03-22 06:11:38 · answer #4 · answered by joshnya68 4 · 0⤊ 0⤋