A 55 Ω and 65 Ω resistor are wired in series in a circuit that contains a 12 V battery. (a) What is the equiva?
A 55 Ω and 65 Ω resistor are wired in series in a circuit that contains a 12 V battery. (a) What is the equivalent resistance of the two resistors? (b) How much current is flowing in the circuit?
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Remember the mnemonic: ParV, SeriQ. Since these resistors are wired in series, R-Equiv is the sum of the two resistors, 55ohms+65ohms, or 120ohms.
Reasoning:
V1+V2= Vtotal, I1=I2 (Series-charge is conserved).
V=IR
I1R1+I2R2=Vtotal
By Kirchhoff's Loop Rule,
I1R1+12R2-12V=0.
I1R1+12R2=12V (Now we see that the total voltage drop must incorporate sum of the 2 resistors)
I(R1+R2)=12V
I=12V/(55+65) ; Total Voltage Drop is 12V, so by V=IR, 12V=12V/(55+65) *Req.... ReQ must equal sum of resistances for voltage drop to be sum of individual voltage drops.
By V=IR, the current is then 12V/120ohms = .1A
In series the resistance is the sum of resistors. To find current take V/R where V is voltage and R is total resistance