A 6 ohm resistor, a 54 ohn resistor, and a 32-Q resistor are connected in series. Calculate their total resistance.
*Can you please show it step by step? oh, and what does "Q" stand for?
2007-03-28 20:40:24 · 4 answers · asked by blah 3 in Science & Mathematics ➔ Physics
Total resistance of N resistors in series:
R(t) = R(1) + R(2) + R(3) + ... + R(n)
R(t) = 6 ohms + 54 ohms + 32 ohms = 92 ohms
The "Q" is actually a capital "Omega" which is the symbol commonly used for the unit of measure "ohms" for resistance (which is futile).
2007-03-28 20:49:34 · answer #1 · answered by Paul McDonald 6 · 1⤊ 1⤋
It is not Q but the greek symbol Omega for Ohm. Since they are connected in series, the total resistance is the sum of the individual resistances. R = R1 + R2 + R3
In this case, R = 6 + 54 + 32 =92 Ohms
2007-03-29 03:48:08 · answer #2 · answered by Swamy 7 · 1⤊ 0⤋
Resistance in series...so R(s) = R1+ R2+ R3
R(s) = 6ohm + 54ohm + 32 ohm
= 60ohm + 32ohm = 92ohm is the ans.
2007-03-29 04:14:59 · answer #3 · answered by Nitya 2 · 0⤊ 0⤋
Are you sure you don't mean omega? In series you just add them together. In parallel you add the inverses and the total is the inverse of the sum.
2007-03-29 03:45:16 · answer #4 · answered by alwaysmoose 7 · 0⤊ 0⤋