Chapter 28: Q.76 (page 794)

Large capacitors can hold a potentially dangerous charge long after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a $120 μ F$ capacitor from a camera flash unit retains a voltage of $150 V$ when an unwary student removes it from the camera. If the student accidentally touches the two terminals with his hands, and if the resistance of his body between his hands is $1.8 k Ω$ , for how long will the current across his chest exceed the danger level of $50 m A$ ?
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 called a $555$ timer. $F I G U R E P 28.77$ shows how the timer’s output pulses, oscillating between $0 V$ and $5 V$ , are controlled with two resistors and a capacitor. The circuit manufacturer tells users that $T_{H}$ , the time the clock output spends in the high $15 V 2$ state,
is $T_{H} = (R_{1} + R_{2}) C \times I N 2 .$ . Similarly, the time spent in the low $10 V 2$ state is $T_{L} = R_{2} C \times I n 2 .$ . You need to design a clock that

Short Answer

Expert verified

$t = 110 m A$ .

Step by step solution

Given information

We need to design a clock.

Simplification

When the switch is off, the capacitor discharge and the current flows through the resistor. To discharge the capacitor, flow through the circuit in time t. The initial current $I_{o}$ is related to the final current I by equation (28.33) in the form

$I = I_{o} e^{- t / R C}$ $(1)$

where R is the resistance and C is the capacitance. Let us solve equation $(1)$ for t to get it by

$t = - R C i n (\frac{I}{I_{o}})$ $(2)$

Ohm's law states that the current $I_{o}$ flows through a resistance due to the potential difference V between resistance

$I_{o} = \frac{∆ V}{R}$ $(3)$

we use the expression of $I_{o}$ from Ohm's law in to equation $(2)$ to the time t in terms of $∆ V$ by

$t = - R C i n (\frac{I}{I_{o}}) = - R C i n (\frac{I}{(∆ V / R)}) = - R C i n (\frac{R 1}{∆ V})$ $(4)$

The final current I represents the danger level current $50 m A$ .Now, we plug the values for R,C,I AND $∆ V$ into equation $(4)$ to get t

$t = - R C i n (\frac{R I}{∆ V}) = - (1.8 \times 10^{3} Ω) (120 \times 10^{- 6} F) I n (\frac{(50 \times 10^{- 3} A) (1.8 \times 10^{3} Ω)}{150 V}) = 0.11 A = 110 m A$ .