(a) [tex]1.11 \Omega[/tex]
When the battery is short-cut, the only resistance in the circuit is the internal resistance of the battery. Therefore, we can apply Ohm's law:
[tex]r=\frac{V}{I}[/tex]
where
V = 20 V is the voltage across the internal resistance of the battery
I = 18 A is the current flowing through it
Solving the equation,
[tex]r=\frac{20 V}{18 A}=1.11\Omega[/tex]
(b) 360 W
The power generated by the battery is given by the equation
[tex]P=VI[/tex]
where
V = 20 V is the voltage of the battery
I = 18 A is the current
Substituting into the formula,
[tex]P=(20 V)(18 A)=360 W[/tex]
(c) 360 J
The energy dissipated by the internal resistance is given by
[tex]E=Pt[/tex]
where
P = 360 W is the power generated
t = 1 s is the time
Solving the equation, we find
[tex]E=(360 W)(1 s)=360 J[/tex]
(d) 1.65 A
The battery is now connected to a [tex]R=11 \Omega[/tex] resistor. This means that the internal resistance of the battery is now connected in series with the other resistor R: so, the total resistance of the circuit is
[tex]R_T = r+R=1.11 \Omega +11 \Omega = 12.11 \Omega[/tex]
And so, the current flowing through the circuit is
[tex]I=\frac{V}{R_T}=\frac{20 V}{12.11\Omega}=1.65 A[/tex]
(e) 29.9 W
The power dissipated in the external resistor is given by
[tex]P=I^2 R[/tex]
where
I = 1.65 A is the current
[tex]R=11 \Omega[/tex] is the resistance
Solving the equation, we find
[tex]P=(1.65 A)^2(11 \Omega)=29.9 W[/tex]
(f) 18.17 V
The terminals of the voltmeter are placed at the two end of the battery. The battery provides an emf of 20 V, however due to the internal resistance, some of this voltage is dropped across the internal resistance. Therefore, the actual potential difference that will be read by the voltmeter will be:
[tex]V=\epsilon - Ir =20 V -(1.65 A)(1.11 \Omega)=18.17 V[/tex]
