The current in R1 equals The potential difference across the terminals of the ba

Question

The current in R1 equals

The potential difference across the terminals of the battery from point "a" to point "b" equals

The current in R2 equals

The voltage across R2 equals

The voltage across R3 equals

The current in R3 equals

The power dissipated in R4 equals

Please show ALL work/explain. Thanks

The circuit below contains a real battery having an EMF of 12.0 V and an internal resistance of 2.00 Ohms The circuit also contains 4 resistors, labeled R, through R4 in the Figure. (12.0 V) R2(4.00 ) R3 (4.00 Q) r (2.00 ) SR4 (8.00 ) R1 (1.00 )

Explanation / Answer

fromt he givendata

   the resultant resistance R is
  

R3,R4 are in series so resultant R of R3,4 is = R3+R4 = 4+8 = 12 ohms

and this is in parallel condition ot R2 so R2,34 = R2*R3,4 /(R2+R3,4) = 4*12/(4+12) = 48/16 = 3 ohms

now this R2,34 is in series to R1 , so R1, R234 = R1+R2334 = 1+3 = 4 ohms

and internal resistance of battery is 2 ohms so total resistance is R = 4+2 = 6 ohms
now current in the circuit is from ohm's law I = 12/6 = 2 A

current passing throught R1 is 2 A

b) p.d across the terminals of the batteryis    V = e- IR = 12 - 2*2 = 8 V

c) current in R2 is = (8/5)4 = 6.4 A
d) voltage across R2 = 6.4*4 = 26.6 V
e) voltage across R3. R4 is equal to voltage across R2 because they are in parallel to each other so p.s is same

f) power dissipation P = i^2 R4 = v^2/R4

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.