(a) The total resistance of the circuit will be 14.0-ohm.
(b) The current in the 6.0-ohm resistor will be 1.5 A.
(c) The potential difference across the battery will be 12-V.
In the series circuit, the amount of current flowing through any component in a series circuit is the same and the sum of the individual resistances equals the overall resistance of any series circuit.
The voltage in a series circuit, the supply voltage, is equal to the total of the individual voltage drops.
(a)
The total resistance of the circuit is found as;
R=R₁+R₂
R=8 -ohm +6-ohm
R=14.0-ohm
(b)
The current in the 6.0-ohm resistor is;
The potential difference across the 6.0-ohm resistor= 12-V.
From Ohm's law;
V=IR
I=V/R
I=12-V/8-ohm
I=1.5-A
c)The potential difference across the battery is;
[tex]\rm I=I_1 \\\\ \frac{V}{R} =\frac{V_1}{R_1} \\\\\frac{V}{14} =\frac{12}{6} \\\\ V=28 \ volt[/tex]
Hence, a) The total resistance of the circuit will be 14.0-ohm.
(b) The current in the 6.0-ohm resistor will be 1.5 A.
(c) The potential difference across the battery will be 12-V.
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