Chap.2 Ex.++ en


Exercises
قريبا إن شاء الله المزيد
Teacher M. Bashier Koukeh
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Exercise 1 — Series Circuit: Addition of Voltages

Physics · Secondary (Première) · Electric circuits

Problem

A circuit consists of a DC generator G, a switch K, a lamp L₁ and two resistors R₁ and R₂, all connected in series. Each voltage is shown as an arrow pointing to the first point of its subscript (for example UAB points to A).

+ − U_PA U_AB U_BC U_CD U_PN K L₁ R₁ R₂ G A B C D P N

Document: series circuit

Given : switch K is closed ; UPN = 12.0 V ; UAB = 5.0 V ; UBC = 4.0 V.
  1. Are the dipoles L₁, R₁ and R₂ connected in series or in parallel? Justify.
  2. Write the law of addition of voltages relating UPN to the voltages across L₁, R₁ and R₂ (K closed).
  3. Calculate UCD.
  4. Which dipole has the greatest voltage across it?
  5. K is now opened. What are UAB, UBC and UCD? Deduce the voltage UPA across the open switch.

Solution

1The three dipoles form a single loop with the generator, with no node: the current has only one path and flows through every dipole. They are therefore connected in series.
2In a series circuit, the voltage across the generator equals the sum of the voltages across the other dipoles: UPN = UAB + UBC + UCD (UPA = 0 since K is closed).
3UCD = UPN − UAB − UBC = 12.0 − 5.0 − 4.0 = 3.0 V
4The lamp L₁: UAB = 5.0 V is the largest of the three.
5K open: the circuit is broken, no current flows, so UAB = UBC = UCD = 0 V. The law of addition then gives UPN = UPA + 0, so UPA = 12.0 V: the whole generator voltage appears across the open switch.

Exercise 2 — Parallel Circuit: Same Voltage Across Branches

Physics · Secondary (Première) · Electric circuits

Problem

A generator G supplies two branches connected in parallel between nodes A and B: branch 1 contains the lamp L₁; branch 2 contains a resistor R and a lamp L₂ in series (middle point C). The generator is considered ideal (its voltage stays constant).

+ − U_PN U_AB U_AC U_CB L₁ R L₂ A B C P N G

Document: parallel circuit

Given : UPN = 6.0 V ; UAC = 2.4 V (voltage across R).
  1. Which dipoles are connected in series? Which dipole is in parallel with the group of the other two?
  2. What is the value of UAB across L₁? State the law used.
  3. Write the law of addition of voltages in branch 2 and calculate UCB.
  4. L₁ is unscrewed. What happens to UAB and UCB? Justify.
  5. L₁ is screwed back and L₂ is unscrewed. Give UAC, UCB and UAB. Check with the law of addition.

Solution

1R and L₂ are in series (same branch, same current). The lamp L₁ is in parallel with the group R + L₂ (both branches are connected to the same nodes A and B).
2Law of uniqueness of voltage: dipoles in parallel between the same nodes have the same voltage across them. Wires connect A and B to the generator terminals, so UAB = UPN = 6.0 V.
3In branch 2: UAB = UAC + UCB ⇒ UCB = 6.0 − 2.4 = 3.6 V
4Branch 2 is still connected to the same nodes A and B, so UAB = 6.0 V does not change. The current in branch 2 is unchanged, so UCB = 3.6 V as well: removing one branch does not affect the other (ideal generator).
5Branch 2 open: no current in R, so UAC = 0 V. The voltage across the L₂ socket is UCB = 6.0 V, and UAB = 6.0 V (L₁ still lit). Check: UAB = UAC + UCB = 0 + 6.0 = 6.0 V.

Exercise 3 — Measuring a Voltage with an Oscilloscope

Physics · Secondary (Première) · Electric circuits

Problem

A DC generator (UPN = 9.0 V) supplies two resistors R₁ and R₂ in series. To measure a voltage, an oscilloscope is connected: its Y terminal is linked to point A and its ground terminal to point N (figure 1). The oscilloscope is in DC coupling; with no voltage, the trace is at the centre of the screen (0 V). Figure 2 shows the screen obtained.

+ − Oscilloscope Y ground R₁ R₂ A N P N G

Figure 1: circuit

y = 2.5 div 0 V (rest) Trace Vertical sensitivity 2.0 V/div DC coupling

Figure 2: oscilloscope screen

Given : vertical sensitivity Sv = 2.0 V/div ; voltage U = n × Sv (n = deflection in divisions).
  1. Which voltage does the oscilloscope measure? Which portion of the circuit is concerned?
  2. Is the oscilloscope connected in series or in parallel with this portion? Why must it not be connected in series?
  3. Read the deflection on figure 2 and calculate the value of the measured voltage.
  4. Deduce the voltage UPA across R₁.
  5. How should the oscilloscope be connected to measure UPA directly? (Where do the Y terminal and the ground go?) Where will the trace be (direction and number of divisions)?
  6. These two connections are swapped (Y at A, ground at P). What happens to the trace? What value is read?

Solution

1Y is at A and the ground at N: the voltage measured is UAN, i.e. the voltage across R₂.
2A voltage is measured between two points: the oscilloscope is connected in parallel across R₂ (Y on one side, ground on the other). In series, it would sit in the loop: its very large input resistance would block the current and disturb the circuit.
3Deflection n = 2.5 div upwards: U = n × Sv = 2.5 × 2.0 = 5.0 V (positive: A is at the higher potential).
4UPN = UPA + UAN ⇒ UPA = 9.0 − 5.0 = 4.0 V
5Y at P, ground at A (in parallel across R₁). The trace rises by n = 4.0 / 2.0 = 2.0 div upwards.
6With Y at A and ground at P, the oscilloscope measures UAP = −UPA. The trace goes down by 2.0 div and the reading is −4.0 V: the sign changes, the absolute value stays 4.0 V.
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