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PHYSICS, M4 EQ-Bank 2

The following circuit has 4 resistors and a 24 V DC supply with a switch.
 

  1. Calculate the value of the single resistor that is equivalent to the 5 \(\Omega\), 20 \(\Omega\), and 40 \(\Omega\) resistors.   (2 marks)

  2. Determine the potential difference across the 15 \(\Omega\) resistor when the switch is closed.   (2 mark)

  3. Determine the current through the 5 \(\Omega\) resistor when the switch is closed.   (2 mark)

Show Answers Only

a.    \(4.6\ \Omega\)

b.    \(18.37\ \text{V}\)

c.   \(1.126\ \text{A}\)

Show Worked Solution
a.     \(R_{series}\) \(= 20 + 40 = 60\ \Omega\)
  \(\dfrac{1}{R_T}\) \(=\dfrac{1}{60}+\dfrac{1}{5}\)
    \(=\dfrac{13}{60}\)
  \(R_T\) \(=\dfrac{60}{13}\)
    \(=4.6\ \Omega\)

 

b.   \(R_{circuit} = 4.6 +15 = 19.6\ \Omega\)

Circuit current:

\(I=\dfrac{V}{R}=\dfrac{24}{19.6}=1.2245\ \text{A}\)
 

Potential difference across the 15 \(\Omega\) resistor:

\(V=IR=1.2245 \times 15=18.37\ \text{V}\)
 

c.   Potential difference across the 5 \(\Omega\) resistor: 

\(V_{(5\ \Omega)}=24-18.37= 5.63\ \text{V}\)
 

Current through the \(5\ \Omega\) resistor:

\(I=\dfrac{V}{R}=\dfrac{5.63}{5}=1.126\ \text{A}\)

PHYSICS, M4 2020 VCE 18a

Students are modelling the effect of the resistance of electrical cables, \(r\), on the transmission of electrical power. They model the cables using the circuit shown in Figure 18.
 

The 24 V DC power supply models the mains power.

Describe the effect of increasing the resistance of the electrical cables, \(r\), on the brightness of the constant resistance globe, \(R\).  (2 marks)

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Increasing the resistance of the electrical cables:

→ Increases the power used by the electrical cables  \(P=IV\). This results in less power in the circuit for the resistance light globe.

→ Therefore, the brightness of the resistance light globe will decrease.

Show Worked Solution

Increasing the resistance of the electrical cables:

→ Increases the power used by the electrical cables  \(P=IV\). This results in less power in the circuit for the resistance light globe.

→ Therefore, the brightness of the resistance light globe will decrease.

PHYSICS, M4 2020 VCE 18

Students are modelling the effect of the resistance of electrical cables, \(r\), on the transmission of electrical power. They model the cables using the circuit shown in Figure 18.
 

The students investigate the effect of changing \(r\) by measuring the current in the electrical cables for a range of values. Their results are shown in Table 1 below.
 

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Resistance of cables,}\ r\ (\Omega) \rule[-1ex]{0pt}{0pt} & \text{Current in cables},\ i\ (\text{A}) & \ \ \ \dfrac{1}{i} \Big{(}\text{A}^{-1}\Big{)}\ \ \  \\
\hline
\rule{0pt}{2.5ex} 2.4 \rule[-1ex]{0pt}{0pt} & 2.4 &  \\
\hline
\rule{0pt}{2.5ex} 3.6 \rule[-1ex]{0pt}{0pt} & 2.0 &  \\
\hline
\rule{0pt}{2.5ex} 6.4 \rule[-1ex]{0pt}{0pt} & 1.7 &  \\
\hline
\rule{0pt}{2.5ex} 7.6 \rule[-1ex]{0pt}{0pt} & 1.5 &  \\
\hline
\rule{0pt}{2.5ex} 10.4 \rule[-1ex]{0pt}{0pt} & 1.3 &  \\
\hline
\end{array}

  1. To analyse the data, the students use the following equation to calculate the resistance of the cables for the circuit.
  2.         \(r=\dfrac{24}{i}-R\)
  3. Show that this equation is true for the circuit shown in Figure 18. Show your working.  (2 marks)

  4. Calculate the values of \(\dfrac{1}{i}\) and write them in the spaces provided in the last column of Table 1 . (1 mark)

  5. Plot a graph of \(r\) on the \(y\)-axis against \(\dfrac{1}{i}\) on the \(x\)-axis on the grid provided below. On your graph:
    • choose an appropriate scale and numbers for the \(x\)-axis
    • draw a straight line of best fit through the plotted points  (3 marks)


  1. Use the straight line of best fit to find the value of the constant resistance globe, \(R\). Give your reasoning.  (2 marks)

Show Answers Only

a.      \(V\) \(=iR\)  
\(V\) \(=i(r+R)\)  
\(\dfrac{V}{i}\) \(=r+R\)  
\(r\) \(=\dfrac{V}{i}-R\)  
\(r\) \(=\dfrac{24}{i}-R\)  

b.   

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Resistance of cables,}\ r\ (\Omega) \rule[-1ex]{0pt}{0pt} & \text{Current in cables},\ i\ (\text{A}) & \ \ \ \dfrac{1}{i} \Big{(}\text{A}^{-1}\Big{)}\ \ \  \\
\hline
\rule{0pt}{2.5ex} 2.4 \rule[-1ex]{0pt}{0pt} & 2.4 & 0.42 \\
\hline
\rule{0pt}{2.5ex} 3.6 \rule[-1ex]{0pt}{0pt} & 2.0 & 0.5 \\
\hline
\rule{0pt}{2.5ex} 6.4 \rule[-1ex]{0pt}{0pt} & 1.7 & 0.59 \\
\hline
\rule{0pt}{2.5ex} 7.6 \rule[-1ex]{0pt}{0pt} & 1.5 & 0.67 \\
\hline
\rule{0pt}{2.5ex} 10.4 \rule[-1ex]{0pt}{0pt} & 1.3 & 0.77 \\
\hline
\end{array}

c.    
       

d.    → The equation for the line:  \(r=\dfrac{24}{i}-R\).

→ The \(y\)-intercept of the graph correlates to the value of \(R\).

→ Reading from the graph, \(R= 7\ \Omega\).

Show Worked Solution

a.      \(V\) \(=iR\)  
\(V\) \(=i(r+R)\)  
\(\dfrac{V}{i}\) \(=r+R\)  
\(r\) \(=\dfrac{V}{i}-R\)  
\(r\) \(=\dfrac{24}{i}-R\)  
♦♦ Mean mark (a) 37%.

b.   

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Resistance of cables,}\ r\ (\Omega) \rule[-1ex]{0pt}{0pt} & \text{Current in cables},\ i\ (\text{A}) & \ \ \ \dfrac{1}{i} \Big{(}\text{A}^{-1}\Big{)}\ \ \  \\
\hline
\rule{0pt}{2.5ex} 2.4 \rule[-1ex]{0pt}{0pt} & 2.4 & 0.42 \\
\hline
\rule{0pt}{2.5ex} 3.6 \rule[-1ex]{0pt}{0pt} & 2.0 & 0.5 \\
\hline
\rule{0pt}{2.5ex} 6.4 \rule[-1ex]{0pt}{0pt} & 1.7 & 0.59 \\
\hline
\rule{0pt}{2.5ex} 7.6 \rule[-1ex]{0pt}{0pt} & 1.5 & 0.67 \\
\hline
\rule{0pt}{2.5ex} 10.4 \rule[-1ex]{0pt}{0pt} & 1.3 & 0.77 \\
\hline
\end{array}

c.    
       

d.    → The equation for the line:  \(r=\dfrac{24}{i}-R\).

→ The \(y\)-intercept of the graph correlates to the value of \(R\).

→ Reading from the graph, \(R= 7\ \Omega\).

♦♦ Mean mark (d) 33%.