Answer:
The current through the cell and voltage across it is 4 A and 2 volts respectively.
Explanation:
The current and voltage across the cell will be calculated through the ohms law. Which is given as,
[tex]V=IR[/tex] (1)
Where,
V=potential difference
I=current through the circuit
R=resistance of the circuit
From the question we have,
R₁=3Ω
R₂=2Ω
R₃=10Ω
R₄=10Ω
The internal resistance of the cell(r)=0.5
The potential difference of the circuit(V)=12volt
The resistors R₁ and R₂ are connected in series. So, their net resistance will be,
[tex]R_s=R_1+R_2[/tex]
[tex]R_s=2+3=5\Omega[/tex] (2)
Now the resistors R₃, R₄ and [tex]R_s[/tex] are connected in parallel. So,
[tex]\frac{1}{R_p}=\frac{1}{R_3} +\frac{1}{R_4} +\frac{1}{R_s}[/tex] (3)
By substituting all the required values in equation (3) we get;
[tex]\frac{1}{R_p}=\frac{1}{10} +\frac{1}{10} +\frac{1}{5}[/tex]
[tex]\frac{1}{R_p}=\frac{1+1+2}{10}[/tex]
[tex]R_p=\frac{10}{4}[/tex]
[tex]R_p=2.5\Omega[/tex] (3)
The net resistance of the circuit is,
[tex]R=R_p+r[/tex] (4)
By substituting the values in equation (4) we get;
[tex]R=2.5+0.5[/tex]
[tex]R=3\Omega[/tex] (5)
By using equation (1) the current through the cell is calculated as,
[tex]12=I\times 3[/tex]
[tex]I=\frac{12}{3}[/tex]
[tex]I=4A[/tex] (6)
The voltage across the cell(e) will also be calculated using equation (1) which is given as,
[tex]e=I\times r[/tex] (7)
By placing the values in equation (7) we get;
[tex]e=4\times 0.5[/tex]
[tex]e=2Volt[/tex]
Hence, the current through the cell and voltage across it is 4 A and 2 volt respectively.
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