The negative terminals of three batteries (2V,4V,6V) are eachconnected to a 20 o

Question

The negative terminals of three batteries (2V,4V,6V) are eachconnected to a 20 ohm resistor, The other sides of each resistorare connected together . the positive terminals are each connectedtogether . How much current flows out of the 4 volt battery?Ifthe wire to the negative terminal of the 6 volt battery breaks sothat no current can flow ,what is the voltage across the gap in thebroken wire?? Which end is at the higher potential??? The negative terminals of three batteries (2V,4V,6V) are eachconnected to a 20 ohm resistor, The other sides of each resistorare connected together . the positive terminals are each connectedtogether . How much current flows out of the 4 volt battery?Ifthe wire to the negative terminal of the 6 volt battery breaks sothat no current can flow ,what is the voltage across the gap in thebroken wire?? Which end is at the higher potential???

Explanation / Answer

the batteries are connected in parallel therefore we get (1/E) = (1/E1) + (1/E2) +(1/E3) E1 = 2 V,E2 = 4 V and E3 = 6V or (1/E) = (1/2) + (1/4) + (1/6) = (11/12) or E = (12/11) V the resistors are connected in parallel therefore we get (1/R) = (1/R1) + (1/R2) +(1/R3) or (1/R) = (1/20) + (1/20) + (1/20) or (1/R) = (3/20) or R = (20/3) the total current flowing through the circuit is E = I * R or I = (E/R) = (12/11) * (3/20) = 0.16 A the total current is I = I1 + I2 + I3 or I2 = I - I1 - I3 or I2 = I - (E1/R) - (E3/R) =0.16 - (2/20) - (6/20) = 0.16 - 0.1 - 0.3 = -0.24 A the negative value of current indicates that the current flowsin the opposite direction the voltage across the gap in the broken wire is (1/E) = (1/E1) + (1/E2) = (1/2) +(1/4) = (2 + 1/4) = (3/4) or E = (4/3) V the right end of the broken wire is at higher potential.Thisis because the right end of the broken wire indicates the positiveterminal of the total emf. or I2 = I - (E1/R) - (E3/R) =0.16 - (2/20) - (6/20) = 0.16 - 0.1 - 0.3 = -0.24 A the negative value of current indicates that the current flowsin the opposite direction the voltage across the gap in the broken wire is (1/E) = (1/E1) + (1/E2) = (1/2) +(1/4) = (2 + 1/4) = (3/4) or E = (4/3) V the right end of the broken wire is at higher potential.Thisis because the right end of the broken wire indicates the positiveterminal of the total emf.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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