2022-PH-FE

#physics

Question 6a

(a) Refer to the circuit shown in Figure 6(a).

i. Calculate the current through the 4 Ω resistor.

ii. Calculate the energy dissipated at the 24 Ω resistor in 12 s.

Solution

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Based on the circuit diagram provided, here is the step-by-step solution.

1. Simplify the Circuit

First, let's find the equivalent resistance of the circuit to determine the total current flowing through the main loop.

Simplify the parallel resistors ( $4 Ω$ and $12 Ω$ ):

R_{p 1} = \frac{4 \times 12}{4 + 12} = \frac{48}{16} = 3 Ω

Find the resistance of the middle branch:

This combined $3 Ω$ resistance is in series with the $5 Ω$ resistor.

R_{middle} = 3 Ω + 5 Ω = 8 Ω

Find the equivalent resistance of the top parallel section:

The top branch ( $24 Ω$ ) is in parallel with the middle branch ( $8 Ω$ ).

R_{parallel} = \frac{24 \times 8}{24 + 8} = \frac{192}{32} = 6 Ω

Find the total resistance ( $R_{total}$ ):

The equivalent resistance of the parallel section ( $6 Ω$ ) is in series with the bottom resistor ( $3 Ω$ ).

R_{total} = 6 Ω + 3 Ω = 9 Ω

2. Calculate Total Current ( $I_{total}$ )

The two voltage sources ( $12 V$ and $3 V$ ) are connected with their positive terminals facing each other (opposing). Therefore, the net voltage ( $V_{net}$ ) is:

V_{net} = 12 V - 3 V = 9 V

Using Ohm's Law ( $I = V / R$ ), the total current flowing from the $12 V$ source is:

I_{total} = \frac{V_{net}}{R_{total}} = \frac{9 V}{9 Ω} = 1 A

Part (i): Calculate the current through the $4 Ω$ resistor

We need to track how the total current of $1 A$ splits through the branches.

Current splitting between the $24 Ω$ branch and the $8 Ω$ middle branch:

Using the Current Divider Rule, the current flowing into the middle branch ( $I_{mid}$ ) is:

I_{mid} = I_{total} \times \frac{R_{top}}{R_{top} + R_{middle}}

I_{mid} = 1 A \times \frac{24}{24 + 8} = \frac{24}{32} = 0.75 A

Current splitting between the $4 Ω$ and $12 Ω$ resistors:

Now, this $0.75 A$ reaches the parallel combination of $4 Ω$ and $12 Ω$ . We apply the Current Divider Rule again to find the current through the $4 Ω$ resistor ( $I_{4 Ω}$ ):

I_{4 Ω} = I_{mid} \times \frac{12}{4 + 12}

I_{4 Ω} = 0.75 A \times \frac{12}{16}

I_{4 Ω} = 0.75 \times 0.75 = 0.5625 A

Answer (i): The current through the $4 Ω$ resistor is $0.5625 A$ .

Part (ii): Calculate the energy dissipated at the $24 Ω$ resistor in $12 s$

Find the current through the $24 Ω$ resistor ( $I_{24}$ ):

Since $0.75 A$ went through the middle branch, the remainder of the total $1 A$ current goes through the top branch.

I_{24} = 1 A - 0.75 A = 0.25 A

Calculate the Power ( $P$ ) dissipated:

P = I^{2} \times R

P = (0.25)^{2} \times 24

P = 0.0625 \times 24 = 1.5 W

Calculate the Energy ( $E$ ):

E = P \times t

E = 1.5 W \times 12 s = 18 J

Answer (ii): The energy dissipated is $18 J$ .

Question 6b

Digital circuits require actions to take place at precise times, so they are controlled by a clock that generates a steady sequence of rectangular voltage pulses. One of the most widely used integrated circuits for creating clock pulses is the 555 timer. Figure 6(b) shows how the timer’s output pulses, oscillating between 0 V and 5 V, are controlled with two resistors and a capacitor.

The manufacturer specifies that the time the clock output spends in the high (5 V) state is
$T_{H} = (R_{1} + R_{2}) C \ln 2$ ,

and the time spent in the low (0 V) state is
$T_{L} = R_{2} C \ln 2$ .

You need to design a clock that oscillates at 10 MHz (i.e., each cycle is $10^{- 7} s$ ) and spends 80 % of each cycle in the high state. Using a $500$ pF capacitor, determine the resistor values.

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Solutions

Based on the problem statement and the circuit formulas provided, here is the step-by-step solution to find the values of $R_{1}$ and $R_{2}$ .

1. Identify the Known Variables

First, we need to break down the frequency and duty cycle into specific time values.

Frequency ( $f$ ): $10 MHz = 10 \times 10^{6} Hz = 10^{7} Hz$
Total Period ( $T$ ): The time for one complete cycle.

T = \frac{1}{f} = \frac{1}{10^{7}} = 10^{- 7} s

Duty Cycle: The system spends 80% of the time in the high state.
- Time High ( $T_{H}$ ): $0.80 \times 10^{- 7} s = 8 \times 10^{- 8} s$
- Time Low ( $T_{L}$ ): The remaining 20%.
  
  $0.20 \times 10^{- 7} s = 2 \times 10^{- 8} s$
Capacitance ( $C$ ): $500 pF = 500 \times 10^{- 12} F = 5 \times 10^{- 10} F$
Constant: $\ln (2) \approx 0.693$

2. Calculate $R_{2}$

We start with $R_{2}$ because the formula for the low time ( $T_{L}$ ) only contains one unknown variable ( $R_{2}$ ).

Formula:

T_{L} = R_{2} C \ln (2)

Rearrange to solve for $R_{2}$ :

R_{2} = \frac{T_{L}}{C \ln (2)}

Substitute the values:

R_{2} = \frac{2 \times 10^{- 8} s}{(5 \times 10^{- 10} F) (0.693)}

R_{2} = \frac{20 \times 10^{- 9}}{0.3465 \times 10^{- 9}}

R_{2} = \frac{20}{0.3465} \approx 57.7 Ω

3. Calculate $R_{1}$

Now that we know $R_{2}$ , we can use the formula for the high time ( $T_{H}$ ) to find $R_{1}$ .

Formula:

T_{H} = (R_{1} + R_{2}) C \ln (2)

Rearrange to solve for $(R_{1} + R_{2})$ :

R_{1} + R_{2} = \frac{T_{H}}{C \ln (2)}

Substitute the values:

R_{1} + R_{2} = \frac{8 \times 10^{- 8} s}{(5 \times 10^{- 10} F) (0.693)}

R_{1} + R_{2} = \frac{80 \times 10^{- 9}}{0.3465 \times 10^{- 9}}

R_{1} + R_{2} = \frac{80}{0.3465} \approx 230.9 Ω

Solve for $R_{1}$ :

R_{1} = 230.9 Ω - R_{2}

R_{1} = 230.9 Ω - 57.7 Ω

R_{1} \approx 173.2 Ω

Final Answer

To achieve a 10 MHz clock with an 80% duty cycle using a 500 pF capacitor:

$R_{1} \approx 173 Ω$
$R_{2} \approx 58 Ω$