The reading of the (ideal) ammeter, in the circuit shown here, equals: (i) I when the key K 7 is closed but the key K 2 is open. (ii) 1 2 when both keys K 1 and K 2 are closed. Find the expression for the resistance of X in terms of the resistances of R and S .

When K is closed and K 2 is open, the current drawn from the battery is I = ξ R + X As S and X are in parallel, total resistance is R ′ = R + S X S + X When both K 1 and K 2 are closed, the current drawn from the battery is I ′ = ξ R ′ = ξ R + S X S + X = ξ ( S + X ) R ( S + X ) + S X The reading of the ammeter will be 1 2 = X S + X I ′ = S S + X × ξ ( S + X ) R ( S + X ) + S X or ξ 2 ( R + X ) = ξ S R S + R X + S X or 2 R S + 2 S X = R S + R X + S X or R S = R X − S X Hence, X = R S R − S .

The reading of the (ideal) ammeter, in the circuit shown here, equals: (i) I when the key K 7 is closed but the key K 2 is open. (ii) 1 2 when both keys K 1 and K 2 are closed. Find the expression for the resistance of X in terms of the resistances of R and S .

When K is closed and K 2 is open, the current drawn from the battery is I = ξ R + X As S and X are in parallel, total resistance is R ′ = R + S X S + X When both K 1 and K 2 are closed, the current drawn from the battery is I ′ = ξ R ′ = ξ R + S X S + X = ξ ( S + X ) R ( S + X ) + S X The reading of the ammeter will be 1 2 = X S + X I ′ = S S + X × ξ ( S + X ) R ( S + X ) + S X or ξ 2 ( R + X ) = ξ S R S + R X + S X or 2 R S + 2 S X = R S + R X + S X or R S = R X − S X Hence, X = R S R − S .

E M B I B E

New Simplified Physics (Vol 1) For Class 12>Chapter 3 - Current Electricity>Problems For Practice>Q 7

EASY

12th CBSE

IMPORTANT

Earn 100

The reading of the (ideal) ammeter, in the circuit shown here, equals:

(i) $I$ when the key $K_{7}$ is closed but the key $K_{2}$ is open.

(ii) $\frac{1}{2}$ when both keys $K_{1}$ and $K_{2}$ are closed.

Find the expression for the resistance of $X$ in terms of the resistances of $R$ and $S$ .

Important Questions on Current Electricity

EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 3 - Current Electricity>Problems For Practice>Q 8

Three resistances, each of

4 Ω

, are connected in the form of an equilateral triangle. Find the effective resistance between its corners.

EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 3 - Current Electricity>Problems For Practice>Q 9

Two resistors are in the ratio

1 : 4

. If these are connected in parallel their total resistance becomes

20 Ω

. Find the value of each resistance.

EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 3 - Current Electricity>Problems For Practice>Q 10

Five resistors are connected as shown in Fig. Find the equivalent resistance between points $B$ and $C$ .

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MEDIUM

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 3 - Current Electricity>Problems For Practice>Q 11

Four resistors of

12 Ω

each are connected in parallel. Three such combinations are then connected in series. What is the total resistance? If a battery of

9 V

emf and negligible internal resistance is connected across the network of resistors, find the current flowing through each resistor.

EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 3 - Current Electricity>Problems For Practice>Q 12

If the reading of the ammeter $A_{1}$ in Fig. is $2.4 A$ , what will the ammeters $A_{2}$ and $A_{3}$ read? Neglect the resistances of the ammeters.

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EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 3 - Current Electricity>Problems For Practice>Q 13

The resistance of the rheostat shown in Fig. is $30 Ω$ . Neglecting the meter resistance, find the minimum and maximum current through the ammeter as the resistance of the rheostat is varied.

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