ohms law for closed circuit

Question 1

diff electromotive force

Answer 1

itis the total work done to transfrer 1C throgh the whole circuit

Question 2

sympolمازا يدل عليع هزا ال

1. Vb
2. r
3. IR
4. Ir

Answer 2

1. it is the electro motive force of the sorcue
2. it is the reistance in side of cell or battery
3. it is the potential difference across the resistance of external circuit OR the terminal voltage across thye source (vt)
4. it is the potential difference across the internal resistance of the source OR drop of voltage duo to the internal resistance

Question 3

what is the rule of ohms law for closed circuit

Answer 3

V_b = IR + Ir

Question 4

when V_b = V_t

Answer 4

when the circuit open ( I = O )

when the internal resistance neglagple (r=0)

Question 5

what is the relation

when the two batteries are connected in same opposite direction and the main battery (high volt ) recharged other battery ( low volt

when the two batteries are connected in same direction the

Answer 5

( VB₁ - VB₂ /R_eq +r₁ +r₂ )

( VB₁+VB₂ /R_eq +r₁ +r₂ )

Question 6

the terminal voltage is always smaller than the e.m.f by the value

……….. and called

Answer 6

Ir which is called drop of voltage

Question 7

experment on ohms law for closed circuit

Answer 7

draw in page 30

To verify Ohm’s law for closed circuit:

1. -Connect circuit as shown.
2. Adjust rheostat to increase the resistance gradually until the current nearly vanishes.
3. Record the reading of the voltmeter.

Observation

• * The terminal voltage is always smaller than the e.m.f by the value [Ir] which is called the drop in voltage.
• * The terminal voltage equals the e.m.f at open circuit when the current nearly vanished. Notes:

Question 8

rule of efficiency of the battery

Answer 8

V/VB x 100

Question 9

if the interial resistancedecrease what will happen to efficiency of the battery and why

Answer 9

increases

That is because the drop in voltage through the battery due to its
internal resistance [Ir] will decrease so the terminal voltage will increase [V=Vb - Ir]

Question 10

The terminal voltage [V] may be greater
than emf ( VB] only ..............

Answer 10

if the battery is being charged

Question 11

when [V = VB Ir]

Answer 11

as the current will be reversed in the battery

Question 12

page 31 الرسمة التحت

Question 13

If you have voltmeter hang on main battery by increasing the rheostat ,what will happen

Answer 13

the current intensity will decrease , and from the relation

V = Vb - Ir the drop of voltage through the main battery decreases So the reading of the voltmeter will increases and tends to reach the value of Vb when the current tends to reach zero

Question 14

If have a voltmeter hang on fixed resistance by increasing the Rheostat

Answer 14

the current intensity will decrease and from relation V = I•R the reading of the voltmeter will decrease and it tends to reach zero when the current tends to reach zero

Question 15

If you have a voltmeter hang on two points not have a resistor that mean and by increasing the rheostat

and why

Answer 15

1. V=zero
2. the reading of the voltmeter remains zero
3. Because there is no decrease in voltage between the 2 point as there is no energy lost to transfer the charges between these 2 points as the resist once of the conducting wires is negligible

Question 16

If you have a voltameter hang on Rheostat V = IR by increasing the rheostat

Answer 16

the current intensity will decrease and from relation reading of voltmeter will increase

for more information page 33 ( IV )

Question 17

If you have voltmeter hang on recharged battery V=Vb + Ir By increasing the rheostat;

Answer 17

the current intensity will decrease, and from the relation

V =Vb + Ir, the reading of the voltmeter will decrease and it tends to reach the value of VB when the current tends to reach zero

NOTE : you see that the V become more tha VB by value Ir when the battery recharged or charged

Question 18

if the electric circuit is switched off . the potential difference between the terminals of the cell = its E .M. F GR

Answer 18

V=VB-IR I = 0 , the drop in voltage (Ir) = 0 V=VB

Question 19

efficiency of battery increases by decrease of internal resistance

Answer 19

because decrease of internal resistan will causes decreasing in drop voltage so the external voltage nearer to the value of E . M . F as (V=VB-Ir) efficiency of battery = V / VB

Question 20

the potential difference between the terminals of the cell is not the same as it E.M.F GR

Answer 20

because work is lost to overcome the internal resistance of the source [r] causes drop in voltage [Ir] so the external voltage is always smaller than E.M.F by the value Ir [V=V_B–Ir]

Question 21

effciency of the battery increases by the decrease of itd internal resistance GR

Answer 21

Decreasing the internal resistance will decreases the drop in voltage so the external voltage will be nearer to the value of the E.M.F AS [V= V_B – Ir]

so the efficiency of the battery will increases

Effciency of the battery = V\V_B