Trial S2 Section B 2019 16 17.docx

SECTION B [15 marks] Answer all questions in this section. 16.

An electric circuit consisting of five resistors and three batteries with negligible internal resistance is shown below. R1= 2.0Ω I1

A

R3= 3.0Ω

I

E2= 4.0V

E1= 8.0V

R5= 4.0Ω

R2= 2.0Ω

B

I3 E3= 10.0V

R4= 3.0Ω

(a) Calculate the currents I1, I2 and I3. [6 marks]

(b) Determine the potential difference VAB.

[2 marks]

17.

The figure shows two long parallel wires P and Q both conducting current I in opposite directions. Point Y is situated at the mid-point between wires P and Q. Points X and Y are perpendicular and equidistant from wire P.

(a) State the ratio of the magnetic flux density at Y to that at X

[3 marks]

(b) If the wires P and Q carry currents of 5 A and 10 A respectively in a vacuum. The force on the wire P is 0.05 N. (i) What is the magnitude and direction of the force on wire Q? [2marks]

(ii) if the current on both wire are the same direction as show on the figure below, sketch a diagram to show the direction exerted by the magnetic flux density, B produced by current I1 on wire Y. On the same diagram, show the direction of the force, F exerted on wire Y due to current I1. [2marks]

Marking Shceme Trial S2 2019 17 An electric circuit consisting of five resistors and three batteries with negligible internal resistance is shown below. R1= 2.0Ω I1

R3= 3.0Ω

A I

E2= 4.0V

E1= 8.0V

R5= 4.0Ω R2= 2.0Ω

B

I3 E3= 10.0V

R4= 3.0Ω

(a) Calculate the currents I1, I2 and I3. [6 marks] I2 = I1 +I3 I3 = I2 - I1 Loop (1): ∑E =∑(IR) 8.0 + 4.0 = I1(2.0)+ I2(4.0) +I1(2.0) 12.0 = 4.0 I1 + 4.0 I2 3.0 = I1 + I2 …………………….(1) Loop(2) : ∑E =∑(IR) 4.0 +10.0 = I2 (4.0)+ I3(3.0) + I3(3.0) 14.0 = 4 I2 + 6 I3 14.0 = 4 I2 + 6 (I2 - I3) 14.0 = 10I2 - 6I1 7.0 = -3I1 + 5I2 …………………(2) (1)x5: (2)-(3):

15.0 = 5I1 + 5I2 -8.0 = -8 I1 I1 = 1.0 A

…………………(3)

Substitute I1 = 1.0 A into (1), 3.0 = 1.0 + I2 I2 = 2.0A Substitute I1 = 1.0 A and I2 = 2.0A into I3 = I2 - I1 = 2.0 -1.0 I3 = 1.0A (b) Determine the potential difference VAB. [2 marks] VAB = E2 - I2 R5 = 4.0 – (2.0)(4.0) = -4.0 V = 4.0 V

17. (a)

1M 1M 1M 1M 1M (b) (i) Currents in opposite directions repel. The forces are of equal magnitude.

(ii)