Series & Parallel Resistors | CAIE 9702 Paper 2 Notes

Paper 2

Physics9702

Series & Parallel Resistors

Components are in series when the same current must pass through each one.

Components are in parallel when both of their ends connect to the same two junctions.

Series: the p.d.s add

Series: one path, same current, p.d.s add up. Parallel: two paths, same p.d., currents split.

The current is the same through every resistor. Since $V=V_1+V_2+\cdots$ and each p.d. is $IR$ , the equivalent resistance is:

R=R_1+R_2+\cdots

Symbols

$R$ = equivalent resistance (\Omega)
$R_1, R_2$ = series resistances (\Omega)

The equivalent series resistance must be larger than every individual resistance.

Parallel: the currents add

Each branch has the same p.d. Since $I=I_1+I_2+\cdots$ and each current is $V/R$ :

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots

Symbols

$R$ = equivalent resistance (\Omega)
$R_1, R_2$ = parallel resistances (\Omega)

A parallel equivalent resistance must be smaller than the smallest branch resistance.
For two resistors only, $R=R_1R_2/(R_1+R_2)$ is a useful shortcut.

Your turn9702/11/M/J/25 Q36

A 100 Ω resistor and a 200 Ω resistor are in series. Their combination is in parallel with a 200 Ω resistor. Calculate the total resistance. [2] (9702/11/M/J/25 Q36)

Show worked answer

The series branch is 300 Ω. Therefore $R=(300\times200)/(300+200)=120\ \Omega$ .

Series & Parallel Resistors

Components are in series when the same current must pass through each one.

Components are in parallel when both of their ends connect to the same two junctions.

Series: the p.d.s add

Series: one path, same current, p.d.s add up. Parallel: two paths, same p.d., currents split.

The current is the same through every resistor. Since $V=V_1+V_2+\cdots$ and each p.d. is $IR$ , the equivalent resistance is:

R=R_1+R_2+\cdots

Symbols

$R$ = equivalent resistance (\Omega)
$R_1, R_2$ = series resistances (\Omega)

The equivalent series resistance must be larger than every individual resistance.

Parallel: the currents add

Each branch has the same p.d. Since $I=I_1+I_2+\cdots$ and each current is $V/R$ :

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots

Symbols

$R$ = equivalent resistance (\Omega)
$R_1, R_2$ = parallel resistances (\Omega)

A parallel equivalent resistance must be smaller than the smallest branch resistance.
For two resistors only, $R=R_1R_2/(R_1+R_2)$ is a useful shortcut.

Your turn9702/11/M/J/25 Q36

A 100 Ω resistor and a 200 Ω resistor are in series. Their combination is in parallel with a 200 Ω resistor. Calculate the total resistance. [2] (9702/11/M/J/25 Q36)

Show worked answer

The series branch is 300 Ω. Therefore $R=(300\times200)/(300+200)=120\ \Omega$ .