1. **Problem statement:** Calculate the number of horizontal divisions occupied by one complete cycle of the waveform displayed on the CRO. 2. **Given:** - Voltage signal: $V(t) = V_0 \sin(2\pi f t)$ with $V_0 = 5$ V and $f = 1$ kHz. - Time-base setting: $1$ ms/div. 3. **Formula:** The period $T$ of the waveform is given by: $$T = \frac{1}{f}$$ 4. **Calculate the period:** $$T = \frac{1}{1000} = 0.001\ \text{s} = 1\ \text{ms}$$ 5. **Calculate the number of horizontal divisions:** Each division corresponds to $1$ ms, so the number of divisions for one cycle is: $$\text{divisions} = \frac{T}{\text{time-base per division}} = \frac{1\ \text{ms}}{1\ \text{ms/div}} = 1\ \text{division}$$ --- 6. **Problem statement:** Derive the expression for the deflection voltage $V_d(t)$ required on the vertical plates to produce a vertical displacement $y(t)$, given the deflection sensitivity $S = 0.5$ mm/V. 7. **Definition:** Deflection sensitivity $S$ is the displacement per unit voltage applied to the deflection plates: $$S = \frac{y}{V_d}$$ 8. **Rearranging for $V_d(t)$:** $$V_d(t) = \frac{y(t)}{S}$$ 9. **Final expression:** $$\boxed{V_d(t) = \frac{y(t)}{0.5}}$$ --- 10. **Problem statement:** Compute the kinetic energy of electrons accelerated by potential $V_a = 2$ kV and explain its effect on brightness. 11. **Given:** - Accelerating potential $V_a = 2000$ V - Electron charge $e = 1.6 \times 10^{-19}$ C 12. **Formula for kinetic energy:** $$KE = e V_a$$ 13. **Calculate kinetic energy:** $$KE = (1.6 \times 10^{-19}) \times 2000 = 3.2 \times 10^{-16}\ \text{J}$$ 14. **Explanation:** Higher kinetic energy means electrons strike the phosphor screen with more energy, causing more intense excitation and thus increasing the brightness of the trace on the CRO screen. **Summary:** - One complete cycle occupies 1 horizontal division. - Deflection voltage $V_d(t) = \frac{y(t)}{0.5}$. - Electron kinetic energy before striking the screen is $3.2 \times 10^{-16}$ J, which increases brightness.